Angular movement problems,

[tanglx61]

Hi people, I am pretty desperate because I have not the least idea of how to do two questions on my homework which is due for tomorrow..Please take a look and if you could do them, that's great, and if you could bother to explain them to me, i will appreciate it a lot!
Thank you! and here are the problems:
1.A uniform, solid disk with mass M and radius R is pivoted about a horizontal axis through its center. A small object of the same mass M is glued to the rim of the disk. If the disk is released from the rest with the small object at the end of a horizontal radius, find hte angular speed when the small object is directly below the axis.2. You are to design a rotating cylindrical axle to life 800-N buckets of cement from the ground to a rooftop 78.0 m above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle, as the axle turns, the bucket will rise.
A) what should the diameter of the axle be in order to raise the buckets at a steady 2.00 cm/s wen it is turning at 7.5 rpm?
B) If instead the axle must give the buckets an upward acceleration of 0.400 m/s^2, what should the angular acceleration of the axle be?

 

[kuruman]

Hi tanglx61, welcome to PF. Please observe the rules of this forum before seeking homework help.

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

NOTE: You MUST show that you have attempted to answer your question in order to receive help. You MUST make use of the homework template, which automatically appears when a new topic is created in the homework help forums.

 

[tomcue]

For the first question, we can use the principle of conservation of angular momentum to solve for the final angular velocity. Initially, the disk and the small object have zero angular momentum as they are at rest. When the disk is released, the small object will start moving in a circular path due to the gravitational force acting on it. At this point, the disk will also start rotating in the opposite direction to conserve angular momentum. We can set up the equation as follows:

Initial angular momentum = Final angular momentum
(0) = (I1 + I2)ωf

Where I1 and I2 are the moments of inertia of the disk and the small object respectively and ωf is the final angular velocity.

We can calculate the moment of inertia of a disk about its center using the formula I = (1/2)MR^2. For the small object, we can approximate it as a point mass rotating around the rim of the disk, so its moment of inertia is MR^2. Substituting these values into the equation, we get:

(0) = ((1/2)MR^2 + MR^2)ωf
ωf = 0

This means that the disk will not have any angular velocity at the moment the small object is directly below the axis. However, the small object will have a linear velocity, which we can calculate using the formula v = ωr, where ω is the angular velocity and r is the radius of the disk. Since the small object is at the end of a radius, its linear velocity will be maximum and equal to v = ωR. Therefore, the maximum speed of the small object will be v = ωR = 0, which means the small object will be stationary at this point.

For the second question, we can use the formula for work done by a torque to solve for the diameter of the axle. The work done by a torque is given by the formula W = τθ, where τ is the torque and θ is the angle through which the torque acts. In this case, the torque is due to the weight of the buckets, so we can write:

W = mgθ

Where m is the mass of the bucket, g is the acceleration due to gravity and θ is the angle through which the bucket has been lifted. We can also express θ in terms of the linear displacement of the bucket using the formula θ = s/r