Asst_31_query

course PHY 201

???????????}assignment #031

U?q?????{???????Physics I

05-03-2008

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21:46:27

experiment to be viewed.

What is the relationship between the angular velocity of the axle around which the string is wound and that of the large disk?

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RESPONSE -->

The angular velocity of each is the same, even the though linear velocity of a given point on each would be different.

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21:46:41

GOOD STUDENT RESPONSE

The angular velocity of the axle and the angular velocity of the disk on the axle would be the same. However, the velocity would be different because they are of different distances from the center. In general, the axle will be moving at a slower speed(velocity) than a point on the outside of the disk. I am not sure if this is what you are asking.

** The speed of the falling object is the same as the speed of a point on the rim of the axle.

The angular velocity of the axle is equal to the speed of a point on its rim divided by its radius: omega = v / r.

The disk rotates with the axle so it has the same angular velocity. **

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21:46:47

If the falling weight accelerates uniformly, does it follow that the rotating disk has a uniform angular acceleration?

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Yes

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21:46:57

GOOD STUDENT RESPONSE yes, because the angle of acceleration is proportional to the velocity of the disk with the radius(which is constant) as the constant of proportionality. And the velocity of the disk will be the same as the velocity of the falling weight which is dependent on the acceleration of the weight.

** If v changes at a uniform rate then since r is uniform, omega = v / r changes at a uniform rate. **

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21:49:25

Principles of Physics and General College Physics Problem 8.28: Moment of inertia of bicycle wheel 66.7 cm diameter, mass 1.25 kg at rim and tire.

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RESPONSE -->

66.7/2 = 33.35 cm = .3335 m

I = 1.25 kg * .3335 m^2 = .139 m N

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21:50:47

The mass of the rim and tire is all located at about the same distance from the axis of rotation, so the rim and tire contribute m * r^2 to the total moment of inertia, where m is the mass and r the distance from the axis of rotation of the rim and tire.

The distance r is half the diameter, or 1/2 * 66.7 cm = 33.4 cm = .334 m, and the mass is given as 1.25 kg, so the moment of inertia of rim and tire is

I = m r^2 = 1.25 kg * (.334 m)^2 = 1.4 kg m^2.

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21:52:11

Why can the mass of the hub be ignored?

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Because it is at the very center of the wheel, thus its r value would be zero (or nearly zero) and so mMassHub*0 = 0

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21:52:24

The radius of the hub is less than 1/5 the radius of the tire; because its moment of inertia is m r^2, where r is its 'average' distance from the axis of rotation, its r^2 will be less than 1/25 as great as for the rim and tire. Even if the mass of the hub is comparable to that of the rim and tire, the 1/25 factor will make its contribution to the moment of inertia pretty much negligible.

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22:00:44

gen Problem 8.38 arm, 3.6 kg ball accel at 7 m/s^2, triceps attachment 2.5 cm below pivot, ball 30 cm above pivot.

Give your solution to the problem.

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RESPONSE -->

I = 3.6 kg * .31 m ^2 = .2883 kg m^2

F = 7.0 m/s^2 * .2883 kg m^2 = 2.02 m N

F would be m a, not I a. Having gotten F you could then multiply by the moment-arm to get the torque, and go from there.

Equivalently you could find the angular acceleration alpha and multiply this by the moment of inertia to get the torque.

You will get the same torque either way.

.025 m * fTricep = 2.02 m N

fTricep = 2.02 m N / .025 m = 80.8 N

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22:08:49

** The moment of inertia of a 3.6 kg ball at a point 30 cm from the axis of rotation is

I = m r^2 = 3.6 kg * (.30 m)^2 = .324 kg m^2.

At a 30 cm distance from axis of rotation the 7 m/s^2 acceleration becomes an angular acceleration of

alpha = a / r = 7 m/s^2 / (.3 m) = 23.3 rad/s^2.

The necessary torque is therefore

tau = I * alpha = .324 kg m^2 * 23.3 rad/s^2 = 7.6 m N, approx..

The muscle exerts its force at a point x = 2.5 cm from the axis of rotation and perpendicular to that axis so we have

F = tau / x = 7.6 m N / (.025 m) = 304 N. **

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RESPONSE -->

I am having a hard time understanding this problem, but I think I have it now. I neglected to convert a to alpha which messed up the final part of the problem. This was tricky.

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22:08:56

Univ. 10.52 (10.44 10th edition). 55 kg wheel .52 m diam ax pressed into wheel 160 N normal force mu =.60. 6.5 m N friction torque; crank handle .5 m long; bring to 120 rev/min in 9 sec; torque required? Force to maintain 120 rev/min? How long to coast to rest if ax removed?

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not univ phy

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22:09:03

** The system is brought from rest to a final angular velocity of 120 rev/min * 1min/60 sec * 2`pi/1 rev = 12.6 rad/s.

The angular acceleration is therefore

alpha = change in omega / change in t = 12.6 rad/s / (9 sec) = 1.4 rad/s^2, approx..

The wheel has moment of inertia I = .5 m r^2 = .5 * 55 kg * (.52 m)^2 = 7.5 kg m^2, approx..

To achieve the necessary angular acceleration we have

tauNet = I * alpha = 7.5 kg m^2 * 1.4 rad/s^2 = 10.5 m N.

The frictional force between ax and wheel is .60 * 160 N = 96 N at the rim of the wheel, resulting in torque

tauFrictAx = -96 N * .52 m = -50 m N.

The frictional torque of the wheel is in the direction opposite motion and is therefore

tauFrict = -6.5 m N.

The net torque is the sum of the torques exerted by the crank and friction:

tauNet = tauFrictAx + tauFrict + tauCrank so that the torque necessary from the crank is

tauCrank = tauNet - tauFrict - tauCrank = 10.5 m N - (-50 m N) - (-6.5 m N) = 67 m N.

The crank is .5 m long; the force necessary to achive the 60.5 m N torque is therefore

F = tau / x = 67 m N / (.5 m) = 134 N.

If the ax is removed then the net torque is just the frictional torque -6.5 m N so angular acceleration is

alpha = -6.5 m N / (7.5 kg m^2) = -.84 rad/s^2 approx.

Starting at 120 rpm = 12.6 rad/s the time to come to rest will be

`dt = `dOmega / alpha = -12.6 rad/s / (-.84 rad/s^2) = 14.5 sec, approx.. **

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This looks good. See my notes. Let me know if you have any questions. &#