#$&*
course phy 201
What would I get on this test? There are certain concept that I need help with but once I understand would I be able to do well?
Time and Date Stamps (logged): 14:09:22 08-01-2011 °³―Έ±±――°±―°°General College Physics (Phy 201) Final Exam
________________________________________
Completely document your work and your reasoning.
You will be graded on your documentation, your reasoning, and the correctness of your conclusions.
________________________________________
Test should be printed using Internet Explorer. If printed from different browser check to be sure test items have not been cut off. If items are cut off then print in Landscape Mode (choose File, Print, click on Properties and check the box next to Landscape, etc.).
Signed by Attendant, with Current Date and Time: ______________________
If picture ID has been matched with student and name as given above, Attendant please sign here: _________
Instructions:
Test is to be taken without reference to text or outside notes.
Graphing Calculator is allowed, as is blank paper or testing center paper.
No time limit but test is to be taken in one sitting.
Please place completed test in Dave Smith's folder, OR mail to Dave Smith, Science and Engineering, Va. Highlands CC, Abingdon, Va., 24212-0828 OR email copy of document to dsmith@vhcc.edu, OR fax to 276-739-2590. Test must be returned by individual or agency supervising test. Test is not to be returned to student after it has been taken. Student may, if proctor deems it feasible, make and retain a copy of the test..
Directions for Student:
Completely document your work.
Numerical answers should be correct to 3 significant figures. You may round off given numerical information to a precision consistent with this standard.
Undocumented and unjustified answers may be counted wrong, and in the case of two-choice or limited-choice answers (e.g., true-false or yes-no) will be counted wrong. Undocumented and unjustified answers, if wrong, never get partial credit. So show your work and explain your reasoning.
Due to a scanner malfunction and other errors some test items may be hard to read, incomplete or even illegible. If this is judged by the instructor to be the case you will not be penalized for these items, but if you complete them and if they help your grade they will be counted. Therefore it is to your advantage to attempt to complete them, if necessary sensibly filling in any questionable parts.
Please write on one side of paper only, and staple test pages together.
Test Problems:
. . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
Problem Number 1
A propeller consists of four blades, each of which can be approximated by a uniform rod 1.6 meters long and having mass 31 kg. To bring the propeller from rest to 800 rpm in 1.4 seconds, how much torque is required? How much energy is required?
.F= ma = 31 kg x 800 rpm / 1.4 sec = 17714.28 N
@& If the initial velocity was 800 m/s then this would be the force on mass 31 kg coming to rest in 1.4 seconds.
However this object is rotating. 800 rpm is an angular velocity, not a linear or translational velocity. That is, 800 rpm is an omega quantity, not a v quantity.
Your instinct to use Newton's Second Law is good. However for rotation, you need to use Newton's Second Law in the form
alpha = tau / I.
You can calculate alpha, the angular acceleration. One way to calculate alpha is
alpha = `dOmega / `dt = 800 rpm / (1.4 sec) = 570 rpm / second.
The convenient unit for angle, however, is not the revolution but the radian. And the standard unit for time is the second, not the minute.
Now 570 rpm / second means
570 (rev / minute) / second.
A revolution is 2 pi radians, and a minute is 60 seconds, so
570 (rev / minute) / second =
570 ( 2 pi radians / (60 seconds) ) / second =
1140 pi / 60 radians / second / second =
58 radians / second^2.
Thus alpha = 58 rad / sec^2.
Now the moment of inertia of a blade is 1/3 m L^2, since each blade spins about its end. There are four blades, so the total moment of inertia is
4 * ( 1/3 M L^2) = 4/3 M L^2 = 4/3 * 31 kg * (1.6 m)^2 = 100 kg m^2, approx.
The required torque is therefore
tau = I * alpha = 100 kg m^2 * 58 rad/s^2 = 5800 kg m^2 / s^2 = 5800 m * N.
*@
Torque = ds xF = 17714.28 N x 1.6 m = 28342.85 mN
@& Torque is moment-arm * force, not displacement * force.*@
.
.work= f x d
.
.
@& Right idea. But in rotation we have
work = tau * `dTheta.
The torque is tau = 5800 m * N.
Starting at 800 rpm and slowing to rest in 1.4 seconds, what is the average angular velocity, and through how many revolutions does the propeller rotate during this time?
How many radians is that?
Your answer is `dTheta.
What therefore is the work?*@
.
.
.
.
.
Problem Number 2
A pendulum of mass 1.5 kg and length 3.44 meters is initially displaced .4472 cm in the horizontal direction from its equilibrium position. The pendulum is released from rest at this position.
How much work does gravity therefore do on the pendulum during its fall to the equilibrium position, and how much work does the pendulum do against gravity?
W = f x d =
F = m x a = 1.5 kg x
it will do the same but will lose some work due to gravity
how do you solve with out time? &&&&&&&
@& The pendulum string will form the hypotenuse of a right triangle, and the .4472 cm displacement will form one of its legs.
What is the length of the other leg?
What is the difference between its length and the length of the string?
How much higher do we conclude the pendulum is, compared to its equilibrium position, because of the displacement?
How much energy does it therefore lose as it falls back to equilibrium?*@
What therefore will be the change in the kinetic energy of the pendulum as it falls?
the PE will change as the pendulum falls
Using energy considerations determine the velocity of pendulum at its equilibrium position.
.how do you solve without time?
.
.
.
@& Set KE equal to the PE loss calculated above.*@
.
.
.
.
.
.
Problem Number 3
A neutron star might have about five times the mass of our Sun, around 10^31 kg, packed into a very nearly perfect sphere of radius roughly 10 km. If you suddenly appeared at the surface of a Neutron star you would almost instantly become a part of that nearly perfect sphere (though you would probably be integrated nearer to your point of contact, you might think of yourself as almost instantly forming a thin coat, like a coat of paint, over the surface of the star).
Suppose you decided to just orbit the star (also fatal, but pretend you somehow managed to maintain physiological and psychological integrity).
If you orbited at a distance of 10.026 km from the center of the star, at what rate would you orbit?
.v^2/r =
v^2/10km = 10.026 km
100
.how do you solve without time ?
.
.
.
@& You have to calculate the orbital velocity be setting the gravitational force on your mass equal to the centripetal force at velocity v. You solve the equation for v and use the result to find the remaining results.
This problem appears with a complete solution in the Introductory Problem Sets.*@
.
.
.
.
.
Problem Number 4
If you have .7 gallons of gasoline, each of which stores approximately 100 megaJoules of chemical energy, which is released by burning the gasoline, then if you could convert all that potential energy to the kinetic energy of a rocket of mass 80 kg, how fast would that rocket be traveling? If you could convert all that energy to gravitational potential energy, how high could you fire the rocket? (Note: Don't ever even think about trying to build a gasoline-powered rocket unless you want to spend the rest of your life in a burn unit, provided you even survive.)
.
..7 gallons
100 megajoules
w= fxd =100,000,000 J = 80 kg x a x d=
@& If you are calculating gravitational potential energy then the basis for calculating the work done against gravity would be a = g = 9.8 m/s^2.*@
how do you solve for acceleration without the time?
f = m x a =
.
.
.
.
.
.
.
.
Problem Number 5
If the coefficient of static friction between tires and pavement is .66, then what torque must be applied to the 66-cm-diameter wheels of an automobile of mass 1450 kg to 'lay rubber', which occurs when friction is broken. code `t
torque = ds x F = .66 m x 1450 kg
F= ma =
1450kg x a =.66
a = 4.55 x 10 ^-4 m/s^2
Im stuck on how to solve the rest
.
.
.
.
.
.
@& If m is the mass of the car, then what is the frictional force corresponding to coefficient of friction .66?
How much torque must be applied to the wheels to exert this much force on the road?*@
.
.
.
.
Problem Number 6
If a simple harmonic oscillator of mass .61 kg is subjected to a restoring force of 3.9 Newtons when displaced .0427 meters from equilibrium, what will be its KE and its PE at equilibrium and halfway to equilibrium if it is released from rest at a displacement of .3904 meters from equilibrium?
. PE = .5 mv^2
will PE and KE be the same?
@& A simple harmonic oscillator at distance x from equilibrium is subject to net force
F_net = - k x.
Its potential energy at position x is 1/2 k x^2.
If F_net = -3.9 Newtons when x = .0427 meters, then what is the value of k?
What therefore is the PE at position x = .39 meters?
What is its PE when it's halfway back to equilibrium?
By how much has its PE changed?
What therefore is the change in its KE?
Setting 1/2 m v^2 = KE, then, what is its speed?
*@
. 61 kg
3.9 N = ma = .61kg x a
a=6.39 m/s^2
@& This is the acceleration at the instant of release.
However it's not helpful in solving this problem. You have to use energy conservation.*@
I dont know what to do next without knowing the time
vf = v0 +adt - do we use that?
@& You don't use this. This applies to uniform acceleration. Net force in this situation changes with position, so the net force is not constant. So acceleration isn't constant.
The equations of uniformly accelerated motion assume constant acceleration, and never apply to simple harmonic motion.*@
.
.
.
.
.
.
.
Problem Number 7
A uniform rod of negligible mass and length 57 cm is constrained to rotate on an axis about its center. Two masses of .756 kg are attached to the rod at a distances of 23.94 cm from the axis of rotation, one on either side of the axis. Two masses of 2.7 kg are attached at the ends of the rods An unknown uniform force is applied to the rod at a position 7.98 cm from the axis of rotation, in the plane of motion of the rod, and at an angle of 32 degrees from the rod. The force is applied as the rod rotates through .12 radians from rest, which requires .9 seconds. The applied force is then removed and, coasting only under the influence of friction, the rod comes to rest after rotating through 3.6 radians, which requires 7 seconds.
Find the net torque for each of the two phases of the motion.
57 cm
2 masses .756 kg with distance 23.94 cm
I = 1/12 M L^2 = 1/12 .753kg x .2394 m = .0431 kg m^2
@& You have the right idea, finding the moment of inertia.
However these masses are not uniformly distributed along a rod, but placed at specific points, so you would calculate the contribution of each mass as m r^2.*@
What is the unknown force?
air resistance
What is the maximum angular momentum of the system? How much of this angular momentum resides in the .756 kg mass?
.momentum is equal to the f = ma?
or f = m v?
.
@& The question asks you to find, not name, the force.
What torque is exerted by an unknown force F applied at 7.98 cm from the axis of rotation, at an angle of 32 degrees from the rod?
How much work does this torque do during the given rotation? At this point your answer will be in terms of F, which remains for the moment unknown.
During the next phase of motion the rod comes to rest while rotating through 3.6 radians in 7 seconds. What therefore is its angular velocity at the beginning of this interval?
What therefore is its KE at the beginning of the interval?
How much KE did it therefore have at the end of the acceleration phase?
Using the moment of inertia, what do you conclude is the net torque exerted by friction?*@
.
.
.
.
.
.
.
.
Problem Number 8
A gun fires a bullet of mass 17 grams out of a barrel 38 cm long. The gun is attached to a spring. From the recoil of the spring and the masses of the gun and the spring we determine that the gun recoiled with a total momentum of 6.6 kg m/s.
With what velocity did the bullet exit the barrel?
f = m x v
@& F = m * a, not m * v.
Momentum is m * v.
*@
6.6 kg m/s = .017 kg x v =
v = 388. 88 m/s is that even possible
@& This is a pretty typical velocity for a bullet fired from a regular gun. A little faster than the speed of sound. Slower than a bullet fired from a high-velocity gun.
*@
Assuming that the bullet accelerated uniformly from rest along the length of the barrel, how long did it take the bullet to accelerate from rest down the length of the barrel?
vf ^2 = v0^2 + 2adt
388. 88 m/s ^2 = 0 + 2 a (.38 m)
a= 510 m/s^2
What was the average force exerted on the bullet as it accelerated along the length of the barrel?
F = ma = 510 m/s^2 x .017 = 8.684 J
@& You didn't put units on .017.
The units would be kg.
The result would therefore be in kg m / s^2.
That's not a Joule. What is it?*@
What average force would be felt by the individual holding the gun for the time the bullet accelerates along the length of the barrel?
.
.the same force but some will be lost due to air resistance
-8.684 J
@& The force is equal and opposite to the force accelerating the bullet, by Newton's Third Law.*@
.
.
.
.
.
.
.
.
Problem Number 9
A car moving at 21 m/s drives over the top of a hill. The top of the hill forms an arc of a vertical circle 130 meters in diameter.
What is the centripetal force holding the car in the circle?
.V^2 / R = 21m/s^2 / 130 m = 3.39 m/s
What therefore is the normal force between the car's tires and the road?
they will cancel each other out, so zero?
@& A car moving along a circular arc experiences a net force equal to the centripetal force.
So the net force on the car is m * 3.39 m/s^2, where m is the mass of the car.
What is the direction of the centripetal force?
What other forces act on the car? You can calculate one of these forces in terms of the mass m of the car; the other is the force exerted by the road. These two forces add up to the net force.
What therefore is the unknown force, and how does this result answer the question?*@
.
Problem Number 10
A simple harmonic oscillator of mass .06 kg is subjected to a net restoring force F = - 80 N/m * x at displacement x from equilibrium.
If the amplitude of its motion is 19.2 meters, what are the maximum magnitudes of its acceleration and velocity?
F = - 80 N/m * x
I dont know where to start
At what displacements from equilibrium can each maximum occur?
.
.
.
.
.
@& You need to thoroughly familiarize yourself with Introductory Problem Set 9.
The key relationships for simple harmonic motion are
F = - k x
PE = 1/2 k x^2
omega = sqrt( k / m ).
You almost always start with at least one of these relationships.
What do you know, and what can you determine from these three relationships?
Now, for the circular model, you have a point moving around a circle of a certain radius at a certain rate.
What is the radius of the circle?
What does omega tell you about the rate?*@
.
.
.
.
.
Problem Number 11
If a simple pendulum of length .98 meters is subjected to a restoring force of 3.6 Newtons when displaced .09506 meters from equilibrium, what is the mass of the pendulum? What will be its KE and its PE at equilibrium and halfway to equilibrium if it is released from rest at a point .09506 meters from equilibrium?
.F = m a
how do you find the mass without time?
.98 m
3.6 N
@& Except for the numbers, this is identical to a previous question.*@
.
Problem Number 12
If a simple harmonic oscillator of mass .27 kg is subjected to a restoring force of 2.3 Newtons when displaced .09628 meters from equilibrium, what will be its KE and its PE at equilibrium and halfway to equilibrium if it is released from rest at a displacement of .1863 meters from equilibrium?
.27 kg
2.3 N
.09628 meters
KE and PE will be the same?
@& This is also identical, except for the numbers, to a previous problem.*@
@& You have the right instincts on many of these problems.
You aren't yet doing very well with simple harmonic motion. You need to learn Introductory Problem Set 9.
With more work over the next few days you should be in good shape to maintain a reasonable grade in the course. You have done an impressive amount of work in the course, and asked many excellent questions. If you can keep this up for a few more days you should end up with a good result.
Answer as many of my questions as you can, and get back to me. I'll be glad to follow up.
*@