Homework 10-31

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course Phy 201

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?

The a_cent is 12.5 m/s/m.

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What centripetal force is required to keep the object moving on its circular path?

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@& You don't appear to have found the 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?

The strap has traveled about a ½ radian. The circumference of the path is about 60cm the strap has to rotate 4 full radians to travel a full rotation.

@& The radian is arc dist / radius, not arc dist / circumference.

Right idea though.

The object has rotated through a little less than 1/2 of a rotation, but that's not 1/2 radian.*@

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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.)

The gravitational force on these two steel balls is about 4.6898*10^-8N.

@& Check this with my calculation. Our results are close, but one of us is off. Might well be me.*@

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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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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.

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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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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What is their total mass M?

There total mass is 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.

@& It looks like you're using 1/2 m r^2 rather than 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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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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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.

The average angular velocity is 5cm/second, 1/12 radians/s,1/12 of a rotation per second.

@& Radians and rotations are different things. Related, but not the same.*@

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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?

The av_ang_accel is -5radians/s^2.

@& Good.*@

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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?

@& You got the moment arm.

See the given solutions for the rest.*@

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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 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.

This is not 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.

This is equilibrium.

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A 5 kg mass and a 6 kg mass suspended from opposite sides of a light frictionless pulley.

This is equilibrium.

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A ball coasting at a constant velocity of 4 m/s.

This is equilibrium.

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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

@& You haven't given your reasoning for these. Some are right and some are wrong, but since you're got a 50-50 chance on each even if your reasoning is wrong, I can't tell if you're right or not.*@

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?

The net torque on that object is 28 kgm^2/radians per second ^2.

@& Units aren't quite right.*@

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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.

@& Revolutions and radians are different.*@

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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?

The work that is done by the torque is 60 kg m^2/s^2.

@& Almost but check the solution. You multiplied by the wrong quantity.*@

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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?

The average net torque is 6 Joules per radian

@& Good, but check the given solution for the meaning of this unit.*@

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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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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?

The angular impulse is 180 kg m ^ 2/ s ^ 2 or 90 kg m^2/s^2 per second.

@& You multiplied by the wrong quantity.*@

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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?

The angular momentum is 24000 kg m^2 * radians/sec.

@& Good. Check for the simplification of this unit.*@

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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?

The force constant is 3/1.

@& The force constant has units.*@

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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?

The rubber band chain will stay in equilibrium.

@& It could but it could also oscillate. Check the given solution.*@

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How long would it take for the reference point to move once around its circle?

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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)?

The PE at that point would be 1.0672*10^-11.

@& This isn't correct. Check the given solution.*@

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If at that point the object is moving at 3 m/s, what its total mechanical energy?

The total mechanical energy is 3.6kg m/s.

@& That's the kinetic energy. The PE you calculated wouldn't add anything significant to this, but the PE of this system is much greater than what you calculated.*@

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@& Here's a copy of the document with my solutions. Compare with yours, especially on the questions where I've indicated that you're on the right track but not quite there, and on the few questions you didn't answer.

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.

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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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`q003: Recall that steel has density around 7.5 grams / cm^3.

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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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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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`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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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

1 radian = 360 deg / (2 pi) = 57 deg, roughly.

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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.

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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.

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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.

`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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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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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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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 velocityis 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.

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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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`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.

#$&*

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.

#$&*

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.

#$&*

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.

#$&*

@& Above is a copy of the document with my solutions. Compare with yours, especially on the questions where I've indicated that you're on the right track but not quite there, and on the few questions you didn't answer.*@

Homework 10-31

@& You don't appear to have found the centripetal force.*@

@& The radian is arc dist / radius, not arc dist / circumference. Right idea though. The object has rotated through a little less than 1/2 of a rotation, but that's not 1/2 radian.*@

@& Check this with my calculation. Our results are close, but one of us is off. Might well be me.*@

@& It looks like you're using 1/2 m r^2 rather than m r^2.*@

@& Radians and rotations are different things. Related, but not the same.*@

@& Good.*@

@& You got the moment arm. See the given solutions for the rest.*@

@& You haven't given your reasoning for these. Some are right and some are wrong, but since you're got a 50-50 chance on each even if your reasoning is wrong, I can't tell if you're right or not.*@

@& Units aren't quite right.*@

@& Revolutions and radians are different.*@

@& Almost but check the solution. You multiplied by the wrong quantity.*@

@& Good, but check the given solution for the meaning of this unit.*@

@& You multiplied by the wrong quantity.*@

@& Good. Check for the simplification of this unit.*@

@& The force constant has units.*@

@& It could but it could also oscillate. Check the given solution.*@

@& This isn't correct. Check the given solution.*@

@& That's the kinetic energy. The PE you calculated wouldn't add anything significant to this, but the PE of this system is much greater than what you calculated.*@

@& Above is a copy of the document with my solutions. Compare with yours, especially on the questions where I've indicated that you're on the right track but not quite there, and on the few questions you didn't answer.*@

@& The radian is arc dist / radius, not arc dist / circumference.

Right idea though.

The object has rotated through a little less than 1/2 of a rotation, but that's not 1/2 radian.*@

@& You got the moment arm.

See the given solutions for the rest.*@