# Angular speed and such

• Nanabit

#### Nanabit

I have a couple problems that I have started but don't quite know where to go with them. If someone could respond ASAP it would be greatly appreciated as one of them will probably appear on a test tomorrow :) Thanks!

1) A potters wheel (a thick stone disk with a radius of .500 m and a mass of 100 kg) is freely rotating at 50.0 rev/min. The potter can stop the wheel in 6.00 sec by pressing a wet rag against the rim and exerting a radial force of 70.0 N. Find the effective coefficient of kinetic friction between the wheel and the rag.

Alright, I know you have to convert 50 rev/min to rev/sec to get omega initial. I also know the sum of torque is Fd or Ialpha. I can figure out both I and alpha. d I'm figuring is the radius. But the problem is, if F is the coefficient times the normal force, I don't even need that? But it doesn't work out anyway?

2) A mass 15.0 kg and a mass 10.0 kg are suspended by a pulley that has a radius 10.0 cm and a mass 3.0 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The masses start from rest a distance 3.00 m apart. Treating the pulley as a uniform disk, determine the speeds of the two masses as they pass each other.

Alright, so once again torque = Ialpha. I can get I. alpha I'm not quite sure .. I played with kinematics but all of them require either time or theta. I don't know where to go when I get torque though.

3) A constant torque of 25.0 Nm is applied to a grindstone whose moment of inertia is .130kgm^2. Using energy principles, find the angular speed after the grindstone has made 15.0 revolutions (neglect friction).

Alright, by using torque = Ialpha I got alpha to be 192 rad/sec^2. However, since I don't know what the initial angular speed was and don't have a time, I couldn't find a kinematic to use.

4) A grinding wheel is in the form of a uniform solid disk having radius of 7.00 cm and a mass of 2.00 kg. It starts from rest and accelerates uniformly under the action of the constant torque of .600 Nm that the motor exerts on the wheel. (a) How long does it take the week to reach its final rotational speed of 1200 rev//min? (b) Through how many revolutions does it turn while accelerating?

Once again, a kinematic problem. I figure you can get I from formulas and thus find alpha.
I = (1/2)MR^2
torque = Ialpha
omegaf = omegai + alpha time
20 = o +(122)t
t = .163 sec

But t = 1.03 sec. If I can get that then I can get part b.

1. a. Use kinematics to find what value of acceleration will slow a wheel down from &omega;o to 0 in 6 seconds. (Convert rev/min to rad/s, not rev/s)
b. Compute the inertia of the wheel. Assume a uniform mass distribution. You should look up the derivation (it's not hard), but you'll find that I = .5*M*r^2
c. The only torque about the center of the wheel is caused by the friction and has a value of f*r, where f = &mu;N, and N is the force applied by the person on the stone. Set this equal to I&alpha; where &alpha; is the acceleration found in part a.

2. &alpha; is the unknown here. Find the sum of torques about the center of the pulley (it will equal the difference in weights times the radius). Take care in establishing a sign convention. Solve for &alpha; using &tau;net = I&alpha; You can then convert &alpha into the linear acceleration of each of the masses (note that the geometric constraint of the rope requires that the velocity of the two masses be equal in magnitude). Knowing the constant acceleration, initial positions, and the fact that the system starts at rest, you should be able to find the answer that is being looked for.

3. Here, you're asked to use energy principles. Assuming the wheel starts at rest, you need to apply the work-energy theorem. Because the applied torque is constant, the work done by this torque is given by &tau;&Delta;&theta;, where &tau; is the torque and &Delta;&theta; is the angle the wheel turns through under that torque. Set the work done equal to the change in kinetic energy, where kinetic energy of an object in pure rotation is given by .5*I*&omega;^2

4. If you convert rev/min into rad/s properly and repeat what you tried to do, you will get the right answer.