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course Phy 201
Im not sure if this sent last night, so I sent it again.
Class 111024'seed' questions ... another resource
A number of students have submitted answers to the questions posed in the 111010 document. This is proving to be a valuable exercise, and it is recommended (but not required) for everyone. If your score on the Major Quiz was less than 93, you should seriously consider submitting your responses. The questions are for the most part relatively short.
You should by this time know the following and be able to apply them to answer questions, set up and analyze lab situations, and solve problems.
How to calculate the units of a product or quotient of quantities having known units.
Definition of average rate
Definition of average velocity
Definition of average acceleration
How to interpret the slope and area of a 'graph trapezoid'.
How to identify an interval
How to declare a positive direction
How to identify the quantities v0, vf, a, `ds, `dt, `dv and vAve for a situation involving uniformly accelerated motion.
The two equations of uniformly accelerated motion that follow from the definitions of average velocity and average acceleration, and how these equations also arise by considering the v0, vf, `dt trapezoid.
How to combine appropriate pairs of the quantities v0, vf, a, `ds, `dt, `dv and vAve using the definitions of average velocity and acceleration and where possible to reason out, over a series of steps, the values of all seven quantities.
How to obtain the third and fourth equations of uniformly accelerated motion from the first two. (Lacking this knowledge, you should at least be able to write down the four equations; however University Physics students should not lack this knowledge).
University Physics students: How to derive the four equations of uniformly accelerated motion using calculus, how to find acceleration, velocity and position functions given one of these functions and, where necessary, intial conditions; how to integrate the force function to find work; how interpretation of a single trapezoid on a graph leads to the interpretation of the derivative and integral of the function represented by that graph.
How to solve any of the four equations of uniformly accelerated motion for any given variable.
How to identify which equation(s) of uniformly accelerated motion can be used with known information to obtain the values of the five quantities v0, vf, `dt, a and `ds.
How to use the equations of uniformly accelerated motoin to find the values of v0, vf, a, `ds and `dt when three of the quantities are known.
How to use Newton's Second Law F_net = m a along with the second and fourth equations of unifomrly accelerated motion to obtain the work-kinetic energy theorem and the impulse-momentum theorem, and the definitions of work, kinetic energy, impulse and momentum.
How to sketch the projection of a vector on a line.
How to calculate the components of a vector on a coordinate plane, given its magnitude and its angle with the positive x axis.
What information is required to add two or more vectors, and how to do so.
How to represent the forces present in a variety of situations, and their components parallel to the coordinate axes.
How to calculate the components of the forces present in a variety of situations, and what information is necessary to do so.
The definition of potential energy.
The difference between work being done on an object and work being done by the object.
The work-energy theorem in terms of `dW_NC, `dKE and `dPE.
To learn to apply these concepts and definitions you need to perform the required lab tasks, answer the questions and problems posed in homework assignments, submit your work on any question or problem you do not completely understand and/or can solve with confidence, and review instructor responses as posted at your access page. A good review of instructor comments will often be accompanied by resubmission of one or more problems.
You should read all of the following problems, make notes on what definitions and concepts you plan to use to solve them, and leave yourself room to later complete the solutions:
General College Physics:
You should read and think about how you would solve all the exercises in Chapters 1-7 which are labeled (I). You should make notes on any of these exercises you are not sure how to do.
Chapter 2 problems 28, 42, 52
Chapter 4 Problems 12, 24, 30, 52
Chapter 6 Problems 6, 12, 22, 42, 52
Chapter 7 Problems 4, 12, 16, 20; 50
Chapter 3, Problems 8, 16, 20, 30
Chapter 5, Problems 8, 14; 30, 36, 48
University Physics:
If you have a lot of trouble with the problems in a chapter then you should work through at least every other odd-numbered exercise (note exercises come before problems). If you have done the questions previously assigned in this course, and read through the text, you should be prepared to do the exercises. If you haven't done the questions previously assigned and can't do most of the exercises, then you really need to work through the previously assigned questions.
Chapter 2, Problems 56, 66, 72, 90
Chapter 3, Problems 44, 48, 56, 64, 68
Chapter 4, Problems 34, 46, 58
Chapter 5, Problems 62, 76, 86, 92, 102, 116
Chapter 6, Problems 56, 58, 64, 74, 82, 84, 90
Chapter 7 Problems 42, 56, 68, 74, 84
Chapter 8 Problems 64, 74, 84, 96, 100, 106
The following definitions and concepts will take us through the rest of the course. These definitions are given along with questions, and answers should be submitted in the usual manner.
Centripetal Acceleration: A point moving on an arc of a circle of radius r with speed v experiences centripetal acceleration a_cent = v^2 / r directed toward the center of the circle.
`q001. What is the centripetal acceleration of a 30 kg mass moving at 10 m/s around a circle of radius 8 m?
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a= v^2/r
a= 10 m/s^2/8 m
100 m/s / 8 m= 12.5 m/s^2
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What centripetal force is required to keep the object moving on its circular path?
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Gravity
@& Multiply centripetal acceleration by mass to get centripetal force.*@
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Radian: A radian is the central angle for which the corresponding arc distance on a circle is equal to the radius of that circle.
`q002: The end of a strap of length 30 cm, rotating about its center, travels 25 cm. Through how many radians has the strap rotated? What is the circumference of the circular path and through how many radians does the strap rotate as it completes one full rotation?
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If the length is 30cm, that means the radius is 15cm, thus making the radian 15cm which means that it only travled through 1, almost 2.
@& RIght idea, but the radius is only 15 cm.*@
Circumfrence = 2pie r
2(3.14)(15)= 94.25 cm which means it can pass through 6.3 radians
@& Good.*@
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Universal Gravitation: Two particles with masses m_1 and m_2, separated by distance r, will each experience a force of attraction to the other. The magnitude of the force of attraction is G m_1 m_2 / r^2 and the direction of the force on one particle is toward the other.
`q003: Recall that steel has density around 7.5 grams / cm^3.
What is the gravitational force between a steel ball of diameter 5 cm and another steel ball of diameter 2.5 cm, if their surfaces are separated by a distance of 1 micron? (Helpful note: As long as they are far enough apart not to touch, they may be regarded as particles, with all the mass of each concentrated at its center. The 1 micron between their surfaces can be neglected for the purpose of finding the force.)
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F= 6.67 *10^-11 (5cm*2.5cm)/ 7.5 g/cm^3 = 1.1 *10^-10
@& Check the given solutions (at the end of this document) on this one. Your numerator should have the product of the masses. Denominator is distance between centers.
Most people didn't get this one so we'll be talking about it tomorrow.*@
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If the gravitational force between these object is the only force they experience, and if they are initially stationary, then what will be the acceleration of each ball? If they were particles where would they meet?
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v=sqrroot GM/r
V= 6.67*10^-11 (5cm)/15cm= 4.7*10^-6
a=v^2/r
a= 4.7*10^-6^2/15cm = 1.5 *10^-22
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How long would it take the spheres, provided they are initially stationary, to move through the 1 micron distance and meet?
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Moment of Inertia: The moment of inertia of a particle of mass m constrained to rotate about a given axis is m r^2, where r is its distance from the axis. The moment of inertia of any rigid object constrained to rotate about a fixed axis is the sum of the m r^2 contributions from all the particles which constitute that object.
Formula: I = sum (m r^2)
Formulas for certain rigid objects with uniform mass distributions (specific conditions and clarifications to be specified later):
hoop: I = M R^2
disk: I = 1/2 M R^2
sphere: I = 2/5 M R^2
rod: I = 1/12 M R^2, I = 1/3 M R^2
`q004: A nut and bolt in the foam disk has mass about 12 grams.
Relative to an axis through the center of the disk, what is the moment of inertia of a bolt located 10 cm from the axis? Answer the same for similar bolts located 4 cm and 7 cm from the axis.
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I= (12g * 10cm^2) = (change the grams into kg) (.12g* 100cm) = 12kg*cm
@& 12 grams is .012 kg, but otherwise good.
SImilar note on the other two.*@
I= (.12kg * 16cm)= 1.92 kg*cm
I= (.12kg * 49cm)= 5.8 kg*cm
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If there are four of each type of bolt, what is their total moment of inertia?
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12*4= 48
1.92*4= 7.68
5.8*4=23.2
Sum of all put together is 78.88 kg*cm
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What is their total mass M?
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12*3= 36*4=144 grams
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If the disk has radius R = 12 cm, then what is the ratio of M R^2 for this disk to the moment of inertia you calculated?
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estimated around 80:144= around 5:9
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Change in angular position: We understand this as the angle through which a rigid object rotates. But motions can be complicated, and rotation of a rigid object is understood relative to an axis, so we often have to be careful about just what we mean by 'the angle through which an object rotates'.
`q005: What is the change in the angular position of a 30 cm strap, rotating about its center, if a point on its end moves 2 cm? Give your answer in radians.
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DTheta= d's/r= 2cm/15cm= .13 radians
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Also reason out,without looking up conversion factors, the number of degrees in this rotation as well as the fraction of a complete rotation. (Hint: If you first figure out the fraction of a complete rotation then it should be easy to figure out the number of degrees. Another hint: How far does that point travel during a complete revolution?) At the very least give this question your best start.
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if it is 30 cm long, that means the circumfrence is going to be 94.2cm which means it is 2cm/94.2 cm which is 1/47.1 or .2
@& That would be .02. The strap rotates through about .13 radians, which is about .02 revolusions.
Except for the misplaced decimal, very good.*@
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More specific definition of change in angular position: The change in the angular position of a rigid object rotating about a fixed axis is equal to the distance moved by one of its particles, divided by the distance of that point from the axis. Alternatively if `r is the vector originating at and perpendicular to the axis and which terminates at a particle not on the axis, then the change in angular position
Angular velocity: The angular velocity omega of a rigid object rotating about a fixed axis is the rate of change of its angular position about that axis, with respect to clock time.
`q006. What is the average angular velocity of a 30 cm strap, rotating about its center, if a point on its end moves 2 cm in .4 second? The units of your answer will include radians. Answer also in units that include degrees, as well as in units that include rotations.
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d'theta/d't= .13 radians/.4 seconds= .325 radians/second
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Angular acceleration: The angular acceleration alpha of a rigid object rotating about a fixed axis is the rate of change of its angular velocity about that axis, with respect to clock time.
`q007: If the strap in the preceding problems slows from angular velocity omega_1 = 5 radians / second to angular velocity omega_2 = 2 radians / second in .6 seconds, what is its average angular acceleration during this interval?
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avg. accel= d'omega/d't= 3 rad/second/ .6 seconds= 5 rad/sec^2
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Line of action of a force: The line of action of a force is the straight line which is parallel to the force, and passes through the point at which the force is applied.
Moment arm: The moment arm of a force about a point is the distance of the line of force from that point. (Refinement: Actually this isn't so; it's the vector from the point to the line, at the closest approach of the line to the point. The distance from the point to the line is the magnitude of the moment arm.)
`q008. One end of a horizontal 8-foot 2 x 4 board rests on a block, the other end in my hand. At a point on the 2 x 4 rests the coupler of a trailer, which exerts a downward force of 500 pounds.
If that point is 1.5 ft. from the end on the block, then what is the magnitude of the moment arm of that force relative to the end of the board on the block?
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8 ft- 1.5 ft= 6.5ft
500lbs *1.5ft= 750 lbs of force
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What is the magnitude of its moment relative to my hand?
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500 lbs * 6.5 ft= 3250 lbs of force
@& The moment arm is from the fulcrum to the point of application; my hand exerts its force at a distance of 8 ft from the point of rotation, not 6.5 ft.
If the board is in equilibrium, then the two moments are equal in magnitude, opposite in direction.
Check the given solution.*@
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Often we consider the moment arm of a force applied to an object which is constrained to rotate about a fixed axis. The moment arm of a force exerted on a rigid object constrained to rotate about a fixed axis is the unique vector originating on the axis and terminating on the line of action of that force, subject to the condition that the vector is perpendicular to both the force and the axis.
Torque: The torque of a force exerted at a point, measured relative to a reference point, is a vector with magnitude and direction. The magnitude of the torque is the product of the magnitudes of the force and the moment arm. The direction is that of the cross product of the moment arm with the force; this direction is perpendicular to both the moment arm and the force, and is visualized using the right-hand rule.
Intuitively the torque exerted by a lever is the product of a force exerted perpendicular to the lever and the distance from the fulcrum at which the force is applied. The distance from the fulcrum is the moment arm of the force.
`q009. One end of an 8-foot 2 x 4 board rests on a block, the other end in my hand. At a point on the 2 x 4 rests the coupler of a trailer, which exerts a downward force of 500 pounds. If that point is 1.5 ft. from the end on the block, then what torque does the force exert about that end?
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the force applied by the trailer is going to be -500 lbs of force,
the force applied by your hand is going to be 3250 lbs of force
the sum of these forces is 2750 lbs of force which is torque
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How much force would I have to exert on my end to equal the magnitude of that torque?
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if it is 6.5 feet away from you then, you will have to only produce a force of around 77 pounds to equal the force from the trailer.
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The torque resulting from a force on a rigid object constrained to rotate about a fixed axis is the cross product of the moment arm of the force, with the force. ... maybe better: A torque is a combination of forces exerted at one or more points of a rigid object which has the effect of changing its angular velocity.
Equilibrium: A system is in equilibrium if the acceleration of its center of mass, and its angular acceleration about any axis, are both zero.
`q010. Which of the following describe equilibrium situations and which do not:?
A block resting on an incline, with the parallel component of its weight equal to 140 Newtons while static friction is capable of producing a force of up to 170 Newtons.
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Does not
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A block sliding along an incline, with the parallel component of its weight equal to 140 Newtons while static friction is capable of producing a force of up to 170 Newtons and kinetic friction a force of up to 140 Newtons.
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equal
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A 5 kg mass and a 6 kg mass suspended from opposite sides of a light frictionless pulley.
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Not equal
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A ball coasting at a constant velocity of 4 m/s.
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equal
@& 3 of your four answers are correct, but with a 50-50 chance on each, I can't tell what you were thinking.
So be sure to check your thinking against the given solutions.*@
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A system is therefore in equilibrium when the net force and net torque on it are both zero.
Formulas: sum(F) = 0, sum(tau) = 0.
sum(F_x) = 0, sum(F_y) = 0, sum(F_z) = 0, sum(tau) = 0
Newton's Second Law for rotating objects: An object whose moment of inertia about an axis is I and which has angular acceleration alpha about that point is experiencing a net torque of I * alpha.
Formula: tau = I * alpha
`q011. What is the net torque on an object whose moment of inertia about a certain point is 4 kg m^2 if it is accelerating at 7 radians / second^2?
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4 kg m^2*7 rad/sec^2= 28 kg m^2 * rad/sec^2
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work\energy in angular motion: The work on a rigid object rotating about a fixed axis, resulting from a torque tau, is `tau dot `d`Theta.
`q012. How much work does a torque of 8 kg m^2 / sec^2 do as the object on which the torque is applied rotates through 6 complete revolutions?
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6.28 radians * 6= 37.68 radians* 8 kg m^2/sec^2= 301 kg m^2/sec^2 *radians
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`q013. What is the work done by a torque of 30 kg m^2 / s^2 when applied for two seconds, during which its angular displacement is 6 radians?
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30 kg m^2/s^2 I= mr^2= 30(6)= 180 kg m^2/s^2 * radians* 2 seconds= 360 kg m^2/s radians
@& Work is still torque * angular displacement. 2 seconds is `dt and is not relevant to this part of the question.*@
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Effect of torque applied through angular displacement
`dW = tau * `dTheta
`q014. If the net torque on a rotating object does 60 Joules of work while the object rotates through 10 radians, what is the average net torque?
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dw/dtheta= tau
60/10= 6 joules per radian
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Effect of torque applied for time interval
tau `dt = `d( I * omega)
for fixed I tau `dt = I * `dOmega
Work-kinetic energy theorem: same as before
`dW_net = `dKE
`q015. By how much will the kinetic energy of a rotating object change as a result of a net torque of 200 kg m^2 / s^2 applied through a rotation of 4 radians?
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tau= 200kg m^2/s^2
d'theta= 4 radians
dw= tau d'theta
dw= 200 kg m^2/s^2 * 4 radians= 800 kg m^2/s^2 * radians
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Effect of a torque: angular impulse
angular impulse = tau_net * `dt
`q016. What is the angular impulse of a torque of 30 kg m^2 / s^2 when applied through an angular displacement of 6 radians, which requires 2 sseconds?
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tau_net= dw/d'theta
180 kg m^2/s^2 * 2 seconds= 360 kg m^2/s
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angular momentum
angular momentum = I * omega
`q017. What is the angular momentum relative to a point of an object whose moment of inertia relative to that point is 1200 kg m^2 and whose angular velocity is 20 rad / sec?
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24000 kg m^2 * rad/sec
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force constant: If the F vs. length graph for the force exerted by an elastic object has constant slope, then that slope is the force constant for that object.
`q018. A graph of the force, in Newtons, exerted by a rubber band chain contains the points (6 cm, 3 Newtons) and (8 cm, 9 Newtons). Assuming that the graph is linear, what is the force constant of this rubber band chain?
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6 cm * 3 N= 18 cm*N
8 cm * 9 N= 72 cm*N
90 cm^2/N^2
@& You got off on the wrong track here. You need to find the slope of the graph between the two points. That's rise / run.
Work it out then check the given solution.*@
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SHM: If the net force on a mass m at position x is - k x, then the mass will either remain in its equilibrium position or oscillate in a manner modeled by the projection on a line through the origin of a reference point moving around a circle with angular frequency omega. If A is the radius of the circle then and amplitude of the motion is A, the speed of the point about the reference circle is v = A * omega, the maximum KE of the object occurs at the equilibrium point and is equal to 1/2 m v^2, the total mechanical energy of the oscillation is 1/2 m v^2, and the potential energy at position x relative to the x = 0 position is 1/2 k x^2.
`q019. A mass of .8 kg is suspended from a rubber band chain having force constant 400 Newtons / meter.
What is its angular frequency omega?
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v= A omega
A= v/omega= 400 N/m / .8= 500 N/m *kg
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How long would it take for the reference point to move once around its circle?
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KE= 1/2mv^2
.4 kg*400^2= 64000 kg
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If at a certain instant the mass is 10 cm from the equilibrium position x = 0, what is the PE at that point (relative to the x = 0 point)?
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If at that point the object is moving at 3 m/s, what its total mechanical energy?
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Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
________________________________________
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@& I didn't give you all the information you need for this one. My fault.
But check the given solution.*@
@&
Check the given solutions below against your solution:
Centripetal Acceleration: A point moving on an arc of a circle of radius r with speed v experiences centripetal acceleration a_cent = v^2 / r directed toward the center of the circle.`q001. What is the centripetal acceleration of a 30 kg mass moving at 10 m/s around a circle of radius 8 m?
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The centripetal acceleration is a_cent = v^2 / r = (10 m/s)^2 / 8 m = 12.5 m/s^2.
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What centripetal force is required to keep the object moving on its circular path?
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The centripetal force is mass * a_cent = 30 kg * 12.5 m/s^2 = 375 N.
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Radian: A radian is the central angle for which the corresponding arc distance on a circle is equal to the radius of that circle.
`q002: The end of a strap of length 30 cm, rotating about its center, travels 25 cm. Through how many radians has the strap rotated? What is the circumference of the circular path and through how many radians does the strap rotate as it completes one full rotation?
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The number of radians is
theta = arc distance / radius = 25 cm / 15 cm.
Note that the radius of the arc is 15 cm, the distance from the center of the strap to the end.
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Universal Gravitation: Two particles with masses m_1 and m_2, separated by distance r, will each experience a force of attraction to the other. The magnitude of the force of attraction is G m_1 m_2 / r^2 and the direction of the force on one particle is toward the other.
`q003: Recall that steel has density around 7.5 grams / cm^3.
What is the gravitational force between a steel ball of diameter 5 cm and another steel ball of diameter 2.5 cm, if their surfaces are separated by a distance of 1 micron? (Helpful note: As long as they are far enough apart not to touch, they may be regarded as particles, with all the mass of each concentrated at its center. The 1 micron between their surfaces can be neglected for the purpose of finding the force.)
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The centers of the balls are separated by the 2.5 cm radius of the first and the 1.25 cm radius of the second. The extra micron is not significant.
The separation is therefore 3.75 cm or .0375 m.
Their volumes are 4 / 3 pi r^3, which gives us volumes of about 60 cm^3 and 8 cm^3.
The masses are therefore about 7.5 g / cm^3 * 60 cm^3 = 450 g and 7.5 g / cm^3 * 8 cm^3 = 56 g, roughly.
The force of attraction is therefore
F = G m1 m2 / r^2 = 6.67 * 10^-11 N m^2 / kg^2 * .45 kg * .056 kg / (.038 m)^2 = 3 * 10^-8 Newtons.
Your numbers, if done accurately, will vary a bit from this result.
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If the gravitational force between these objects is the only force they experience, and if they are initially stationary, then what will be the acceleration of each ball? If they were particles where would they meet?
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The acceleration of each is calculated as a = F_net / m.
For the first ball this is a = 3 * 10^-8 N / (.45 kg) = 6.7 * 10^-8 m/s^2.
For the second we get a = 3 * 10^-8 N / (.056 kg) = 6 * 10^-7 m/s^2.
Calculations are not precise but are reasonably close to correct values.
Calculated precisely, the less massive object has 1/8 the mass of the greater, and 8 times the acceleration.
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How long would it take the spheres, provided they are initially stationary, to move through the 1 micron distance and meet?
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The spheres would both accelerate from rest at the calculated rates, until the sum of the distances they have covered is 1 micron.
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Moment of Inertia: The moment of inertia of a particle of mass m constrained to rotate about a given axis is m r^2, where r is its distance from the axis. The moment of inertia of any rigid object constrained to rotate about a fixed axis is the sum of the m r^2 contributions from all the particles which constitute that object.
Formula: I = sum (m r^2)
Formulas for certain rigid objects with uniform mass distributions (specific conditions and clarifications to be specified later):
hoop: I = M R^2
disk: I = 1/2 M R^2
sphere: I = 2/5 M R^2
rod: I = 1/12 M R^2, I = 1/3 M R^2
`q004: A nut and bolt in the foam disk has mass about 12 grams.
Relative to an axis through the center of the disk, what is the moment of inertia of a bolt located 10 cm from the axis? Answer the same for similar bolts located 4 cm and 7 cm from the axis.
For the nut and bolt 10cm from the center the moment of inertia is 600g*cm, at 4cm from the axis it is 96g*cm, and 7cm from the axis it is 294g*cm.
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All the mass of a bolt is concentrated close to the a single distance from the axis of rotation. Each bolt therefore has moment of inertia m r^2.
For each bolt at 10 cm we have m r^2 = 12 grams * (10 cm)^2 = 1200 g cm^2.
At 7 cm we have m r^2 = 12 g * (7 cm)^2 = 590 g cm^2.
At 4 cm we have m r^2 = 12 g * (4 cm)^2 = 190 g cm^2.
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If there are four of each type of bolt, what is their total moment of inertia?
The total moment of inertia is 3960g*cm.
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Multiplying each moment of inertia by 4 and adding the results we get a total moment of inertia close to 8000 g cm^2.
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What is their total mass M?
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Multiplying the mass of a single bolt by 12 and we get a total mass of 144 grams.
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If the disk has radius R = 12 cm, then what is the ratio of M R^2 for this disk to the moment of inertia you calculated?
The disks moment of inertia would be 10368g*cm.
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M = 144 grams and R = 12 cm we get M R^2 = 144 g * (12 cm)^2 = 20 000 g cm^2, roughly. This would be the moment of inertia if all 12 bolts were 12 cm from the axis.
So M R^2 is about 20 000 g cm^2 / (8000 g cm^2) = 2.5 times as great as the actual moment of inertia of the bolts.
The actual moment of inertia is less because the bolts are concentrated closer to the center than the 12 cm assumed in the calculation M R^2.
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Change in angular position: We understand this as the angle through which a rigid object rotates. But motions can be complicated, and rotation of a rigid object is understood relative to an axis, so we often have to be careful about just what we mean by 'the angle through which an object rotates'.
`q005: What is the change in the angular position of a 30 cm strap, rotating about its center, if a point on its end moves 2 cm? Give your answer in radians.
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The radius of the motion of an endpoint is 15 cm, so if an endpoint moves 2 cm the angular position changes by
angle = arc dist / radius = 2 cm / (15 cm) = .13,
which means .13 radians.
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Also reason out,without looking up conversion factors, the number of degrees in this rotation as well as the fraction of a complete rotation. (Hint: If you first figure out the fraction of a complete rotation then it should be easy to figure out the number of degrees. Another hint: How far does that point travel during a complete revolution?) At the very least give this question your best start.
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The circumference of a circle is 2 pi r.
So the angle corresponding to a complete circuit around the circle is
angle = arc dist / radius = 2 pi r / r = 2 pi.
A complete circle corresponds to an angular displacement of 2 pi radians.
2 pi is about 6 so a complete circle corresponds to about 6 radians.
A complete circle also corresponds to an angular displacement of 360 degrees. So about 6 radians corresponds to 360 deg, and we conclude that 1 radian is about 360 / 6 deg = 60 deg. This is only an estimate.
More accurately
2 pi radians = 360 degrees
so
1 radian = 360 deg / (2 pi) = 57 deg, roughly.
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More specific definition of change in angular position: The change in the angular position of a rigid object rotating about a fixed axis is equal to the distance moved by one of its particles, divided by the distance of that point from the axis. Alternatively if `r is the vector originating at and perpendicular to the axis and which terminates at a particle not on the axis, then the change in angular position
Angular velocity: The angular velocity omega of a rigid object rotating about a fixed axis is the rate of change of its angular position about that axis, with respect to clock time.
`q006. What is the average angular velocity of a 30 cm strap, rotating about its center, if a point on its end moves 2 cm in .4 second? The units of your answer will include radians. Answer also in units that include degrees, as well as in units that include rotations.
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As seen earlier the given arc displacement of the end of this strap corresponds to an angle of about .13 radian.
This occurs in .4 second.
The average angular velocity is therefore
omega_ave = ave rate of change of angular position with respect to clock time = (change in angular position) / (change in clock time) = .13 rad / (.4 sec) = .33 rad / sec, approx.
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Angular acceleration: The angular acceleration alpha of a rigid object rotating about a fixed axis is the rate of change of its angular velocity about that axis, with respect to clock time.
`q007: If the strap in the preceding problems slows from angular velocity omega_1 = 5 radians / second to angular velocity omega_2 = 2 radians / second in .6 seconds, what is its average angular acceleration during this interval?
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The average angular acceleration is ave rate of change of angular velocity with respect to clock time, so
alpha_ave = (change in angular vel) / (change in clock time) = (2 rad/s - 5 rad/s) / (.6 sec) = (-3 rad/s) / (.6 s) = -5 rad/s^2.
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Line of action of a force: The line of action of a force is the straight line which is parallel to the force, and passes through the point at which the force is applied.
Moment arm: The moment arm of a force about a point is the distance of the line of force from that point. (Refinement: Actually this isn't so; it's the vector from the point to the line, at the closest approach of the line to the point. The distance from the point to the line is the magnitude of the moment arm.)
`q008. One end of a horizontal 8-foot 2 x 4 board rests on a block, the other end in my hand. At a point on the 2 x 4 rests the coupler of a trailer, which exerts a downward force of 500 pounds.
If that point is 1.5 ft. from the end on the block, then what is the magnitude of the moment arm of that force relative to the end of the board on the block?
The moment arm of that force is 1.5ft.
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What is the magnitude of its moment relative to my hand?
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Assume that the board is horizontal. Then the line of action of the 500 pound force is perpendicular to the board, and passes 1.5 ft from the fulcrum. The moment arm is therefore 1.5 ft.
The force is 500 lb, so the moment is 500 lb * 1.5 ft = 750 lb * ft.
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Often we consider the moment arm of a force applied to an object which is constrained to rotate about a fixed axis. The moment arm of a force exerted on a rigid object constrained to rotate about a fixed axis is the unique vector originating on the axis and terminating on the line of action of that force, subject to the condition that the vector is perpendicular to both the force and the axis.
Torque: The torque of a force exerted at a point, measured relative to a reference point, is a vector with magnitude and direction. The magnitude of the torque is the product of the magnitudes of the force and the moment arm. The direction is that of the cross product of the moment arm with the force; this direction is perpendicular to both the moment arm and the force, and is visualized using the right-hand rule.
Intuitively the torque exerted by a lever is the product of a force exerted perpendicular to the lever and the distance from the fulcrum at which the force is applied. The distance from the fulcrum is the moment arm of the force.
`q009. One end of an 8-foot 2 x 4 board rests on a block, the other end in my hand. At a point on the 2 x 4 rests the coupler of a trailer, which exerts a downward force of 500 pounds. If that point is 1.5 ft. from the end on the block, then what torque does the force exert about that end?
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How much force would I have to exert on my end to equal the magnitude of that torque?
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The torque tending to rotate the board 'down' is 750 lb * ft, so I have to exert a torque of 750 lb * ft tending to rotate the board 'up'.
My hand is 8 ft from the fulcrum, so the upward force I exert has a line of force passing 8 ft from the fulcrum.
If the force I exert is F, then, my torque is F * 8 ft.
Thus
F * 8 ft = 750 lb * ft
and
F = 750 lb * ft / (8 ft) = 90 lb, approx.
Another less formal way of getting the idea:
My force is 8 / 1.5 times as far from the fulcrum so I have 8 / 1.5 times the leverage, or 5.3 times the leverage.
So I require a force 5.3 times smaller, or 500 lb / (5.3) = 90 lb, approx.
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The torque resulting from a force on a rigid object constrained to rotate about a fixed axis is the cross product of the moment arm of the force, with the force. ... maybe better: A torque is a combination of forces exerted at one or more points of a rigid object which has the effect of changing its angular velocity.
Equilibrium: A system is in equilibrium if the acceleration of its center of mass, and its angular acceleration about any axis, are both zero.
`q010. Which of the following describe equilibrium situations and which do not:?
A block resting on an incline, with the parallel component of its weight equal to 140 Newtons while static friction is capable of producing a force of up to 170 Newtons.
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Friction will oppose motion down the incline.
It only takes 140 N in this direction to keep the block in equilibrium.
Friction can exert up to 170 N in this direction.
So friction can exert the necessary 140 N, which it does, keeping the block in equilibrium.
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A block sliding along an incline, with the parallel component of its weight equal to 140 Newtons while static friction is capable of producing a force of up to 170 Newtons and kinetic friction a force of up to 140 Newtons.
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The block is sliding so it's the kinetic friction that applies.
Kinetic friction is capable of producing 140 N to oppose the 140 N component of the weight along the incline.
So this is equilibrium.
The net force is zero, the acceleration is zero, so the object slides at constant velocity.
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A 5 kg mass and a 6 kg mass suspended from opposite sides of a light frictionless pulley.
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Gravity exerts unequal forces on the unequal masses.
The forces tend to accelerate the system in opposite directions. If the gravitational forces were equal this would be an equilibrium situation, with net force equal to zero.
However the gravitational forces are not equal, so the net force is not zero, and the system is not in equilibrium.
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A ball coasting at a constant velocity of 4 m/s.
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The acceleration of an object moving at constant velocity is zero, so this is an equilibrium situation.
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A system is therefore in equilibrium when the net force and net torque on it are both zero.
Formulas: sum(F) = 0, sum(tau) = 0.
sum(F_x) = 0, sum(F_y) = 0, sum(F_z) = 0, sum(tau) = 0
Newton's Second Law for rotating objects: An object whose moment of inertia about an axis is I and which has angular acceleration alpha about that point is experiencing a net torque of I * alpha.
Formula: tau = I * alpha
`q011. What is the net torque on an object whose moment of inertia about a certain point is 4 kg m^2 if it is accelerating at 7 radians / second^2?
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The net torque is
tau_net = I * alpha = 4 kg m^2 * 7 rad/s^2 = 28 kg m^2 / s^2.
Note that since a radian is an arc distance divided by a radius, an arc distance is the angle in radians, multiplied by the radius. So m * rad can result in just plain m.
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work\energy in angular motion: The work on a rigid object rotating about a fixed axis, resulting from a torque tau, is `tau dot `d`Theta.
`q012. How much work does a torque of 8 kg m^2 / sec^2 do as the object on which the torque is applied rotates through 6 complete revolutions?
The torque does a work of 48kg m^2/sec^2.
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The work is
`dW = tau * `dTheta = 8 kg m^2 / s^2 * 6 revolutions = 48 kg m^2 * revolution / s^2.
A revolution corresponds to 2 pi radians of angular displacement, so
`dW = 48 kg m^2 * (2 pi rad) / s^2 = 96 pi kg m^2 / s^2, or about 300 kg m^2 / s^2.
A kg m^2 / s^2 is a Joule, so this is about 300 Joules.
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`q013. What is the work done by a torque of 30 kg m^2 / s^2 when applied for two seconds, during which its angular displacement is 6 radians?
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Work is torque * angular displacement so
`dW = tau * `dTheta = 30 kg m^2 / s^2 * 6 rad = 180 kg m^2 / s^2, or 180 Joules.
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Effect of torque applied through angular displacement
`dW = tau * `dTheta
`q014. If the net torque on a rotating object does 60 Joules of work while the object rotates through 10 radians, what is the average net torque?
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`dW = tau_ave * `dTheta, so
tau_ave = `dW / `dTheta = 60 J / (10 rad) = (60 kg m^2 / s^2) / (10 rad) = 60 kg m^2 / s^2.
The fundamental units of the net torque are the same as those of work.
60 kg m^2 / s^2 = 60 N * m.
Since in this case we are talking about torque, we write the unit as m * N, in order to distinguish that this is a torque.
Our average torque is thus
tau_ave = 60 m * N.
We would not express our net torque as 60 Joules. It's a torque, not an amount of work.
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Effect of torque applied for time interval
tau `dt = `d( I * omega)
for fixed I tau `dt = I * `dOmega
Work-kinetic energy theorem: same as before
`dW_net = `dKE
`q015. By how much will the kinetic energy of a rotating object change as a result of a net torque of 200 kg m^2 / s^2 applied through a rotation of 4 radians?
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`dW_net = tau_net * `dTheta = 200 kg m^2 / s^2 * 4 rad = 800 kg m^2 / s^2, or 800 Joules.
So the KE of the object increases by 800 Joules.
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Effect of a torque: angular impulse
angular impulse = tau_net * `dt
`q016. What is the angular impulse of a torque of 30 kg m^2 / s^2 when applied through an angular displacement of 6 radians, which requires 2 seconds?
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angular impulse = tau_net * `dt = 30 kg m^2 / s^2 * 2 sec = 60 kg m^2 / s.
This is not a unit of work or a unit of torque, it is the unit of angular impulse.
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angular momentum
angular momentum = I * omega
`q017. What is the angular momentum relative to a point of an object whose moment of inertia relative to that point is 1200 kg m^2 and whose angular velocity is 20 rad / sec?
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The angular momentum is
angular momentum = I * omega = 1200 kg m^2 * 20 rad/sec = 24 000 kg m^2 / s.
Note that the unit is identical to that of angular impulse.
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force constant: If the F vs. length graph for the force exerted by an elastic object has constant slope, then that slope is the force constant for that object.
`q018. A graph of the force, in Newtons, exerted by a rubber band chain contains the points (6 cm, 3 Newtons) and (8 cm, 9 Newtons). Assuming that the graph is linear, what is the force constant of this rubber band chain?
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The slope of the graph is
slope = rise / run = (9 N - 3 N) / (8 cm - 6 cm) = 6 N / (2 cm) = 3 N / cm.
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SHM: If the net force on a mass m at position x is - k x, then the mass will either remain in its equilibrium position or oscillate in a manner modeled by the projection on a line through the origin of a reference point moving around a circle with angular frequency omega. If A is the radius of the circle then and amplitude of the motion is A, the speed of the point about the reference circle is v = A * omega, the maximum KE of the object occurs at the equilibrium point and is equal to 1/2 m v^2, the total mechanical energy of the oscillation is 1/2 m v^2, and the potential energy at position x relative to the x = 0 position is 1/2 k x^2.
`q019. A mass of .8 kg is suspended from a rubber band chain having force constant 400 Newtons / meter.
What is its angular frequency omega?
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The system may remain in equilibrium. If it is disturbed from its equilibrium point it will oscillate with angular frequency
omega = sqrt(k / m) = sqrt( (400 N / m) / (.8 kg) ) = sqrt( 500 / s^2) = 22 radians / s.
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How long would it take for the reference point to move once around its circle?
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To move once around the circle the angular displacement would have to be 2 pi radians.
At 22 rad / s this would require time
`dt = 2 pi rad / (22 rad/s) = .3 second, approximately.
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If at a certain instant the mass is 10 cm from the equilibrium position x = 0, what is the PE at that point (relative to the x = 0 point)?
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The PE would be
PE = 1/2 k x^2 = 1/2 * 400 N / m * (.1 m)^2 = 2 N * m = 2 Joules.
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If at that point the object is moving at 3 m/s, what its total mechanical energy?
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The total mechanical energy is PE + KE, so we get
total mechanical energy = 1/2 k x^2 + 1/2 m v^2 = 1/2 ( 400 N/m) * (.1 m)^2 + 1/2 * .8 kg * (3 m/s)^2 = 2 J + 3.6 J = 5.6 J.
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You're doing very well here. A few glitches, and you were off track on at least one, but you got most of these questions.
Solutions are given in the document directly above this note. Check my inserted notes, and check your solutions against the given solutions.
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