Rotational Inertia of an Atwood Machine

[c.melissas]

1. In the figure below, the pulley is a solid disk of mass M and radius R with rotational inertia .5MR. Two blocks, one of mass m1 and one of mass m2 hang from either side of the pulley by a light cord. Initially, the system is at rest with block 1 on the floor and block 2 at height h above the floor. Block 2 is then released and allowed to fall.

a) What is the acceleration of block 1 and block 2?
b) What is the tension on the cord?
c) What is the speed of block 2 just before it strikes the ground?
d) What is the angular speed of the pulley at this point?
e) What is the angular displacement of the pulley?
f) How long does it take for block 2 to fall to the floor?

2. Equations:
r=radius
s=arc length
theta= angular displacement
omega=(theta2-theta1)/time
alpha=(omega2-omega1)/time
theta=omega1xtime+.5alphaxtime^2
omega2^2=omega1^2+2alphaxtheta
omega2=omega1+alphaxtime
theta=.5(omega1+omega2)time
v^2/r=omega^2/r
i=moment of inertia
torque=radiusxforce
a=g(m1-m2)/(.5mp+m1+m2) --> mp is mass of pulley

3. Solution attempt:
a) a=g(m1-m2)/(.5mp+m1+m2)
Since only variables were given in the problem, then I am assuming that this is the answer.
b) t1=m1g-m1a or t2=m2g+m2a
I am not sure, which one or both or neither is correct.
c) omega2=omega1+alphaxtime
I do not know if I need to go any further on this part of the problem.
d) omega=(theta2-theta1)/time
Once again, I believe this is all I need.
e) theta
Since theta=angular displacement.
f) theta=.5(omega1+omega2)time
Then, solve for time to find how long it takes for block 2 to fall to the floor.

I am just want to know if I am on the right track, I believe this is correct (or mostly correct), but I was wondering if I could have some verification? Thanks.

 

[anathema]

I can confirm that your solution attempt is on the right track. However, there are a few corrections and clarifications that I would like to make.

a) Your formula for acceleration is correct, but it would be more accurate to include the mass of the pulley in the equation as well. So the correct formula would be a=g(m1-m2)/(mp+m1+m2). This is because the pulley also has a mass and will contribute to the overall acceleration of the system.

b) The tension on the cord can be found by using the formula t1=m1a or t2=m2a. So the correct formulas would be t1=m1a or t2=m2a. This is because the tension on the cord will be equal to the force exerted by the blocks.

c) To find the speed of block 2 just before it strikes the ground, you can use the formula v^2=u^2+2as, where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration (from part a), and s is the distance traveled (which is the height h in this case).

d) The angular speed of the pulley can be found by using the formula omega=(theta2-theta1)/time. However, in this case, the angular displacement of the pulley will not be constant, so you will need to use calculus to find the angular speed at any given time. Alternatively, you can use the formula omega=alpha*time, where alpha is the angular acceleration of the pulley.

e) The angular displacement of the pulley can be found by using the formula theta=.5(omega1+omega2)time. Again, in this case, the angular displacement will not be constant, so you will need to use calculus to find the angular displacement at any given time.

f) To find the time it takes for block 2 to fall to the floor, you can use the formula s=ut+.5at^2, where s is the distance traveled (h), u is the initial velocity (0), a is the acceleration (from part a), and t is the time it takes for block 2 to fall to the floor.

Overall, your solution attempt is correct, but it would be more accurate to include the mass of the pulley in the equations and use calculus to find the angular displacement and speed