Atwood Machine, Rotational Inertia, and Energy

[rvhockey]

In the figure below, the pulley is a solid disk of mass M and radius R, with rotational inertia (MR²)/2. 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 held at height h above the floor. Block 2 is then released and allowed to fall.
a. What is the speed of Block 2 just before it strikes the ground?
b. What is the angular speed of the pulley at this moment?
c. What's the angular displacement of the pulley?
d. How long does it take for Block 2 to fall to the floor?



(MR²)/2 = I
mgh=GPE
(mv²)/2 =KE_trans
Iw(omega)²=KE_rot



I can't figure out how to do it, but the answers are
a. v = sqrt((2gh(m₂-m₁))/(m₁+m₂+(M/2)))
b. w(omega) = (1/R) * sqrt((2gh(m₂-m₁))/(m₁+m₂+(M/2)))
c. ??
d. t = h * sqrt((2m₁+2m₂+M)/(gh(m₂-m₁)))

 

[heth]

Think about it in terms of energy conservation. Do a sketch of the system at the start and at the end of the period you're interested in.

Then play spot the difference. Where has energy been transferred?

 

[thomate1]

I would approach this problem by first understanding the concepts involved: Atwood machine, rotational inertia, and energy.

An Atwood machine is a device used to demonstrate the principles of mechanical energy conservation. It consists of two masses connected by a string or cord that passes over a pulley. In this case, the pulley is a solid disk with rotational inertia (MR2)/2.

Rotational inertia is the resistance of an object to changes in its rotational motion. It depends on the mass and distribution of the object's mass around its axis of rotation. In this case, the rotational inertia of the pulley is (MR2)/2.

Energy is the ability to do work. In this case, there are two types of energy involved: gravitational potential energy (GPE) and kinetic energy (KE). GPE is the energy an object possesses due to its position in a gravitational field, and it is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. KE is the energy an object possesses due to its motion, and it is given by (mv2)/2, where m is the mass and v is the speed.

Now, let's answer the questions:

a. The speed of Block 2 just before it strikes the ground can be calculated using the conservation of mechanical energy equation: mgh = (mv2)/2. Rearranging the equation, we get v = sqrt(2gh). However, we need to take into account the masses on either side of the pulley, so the equation becomes v = sqrt((2gh(m2-m1))/(m1+m2+(M/2))). This is the speed of Block 2 just before it strikes the ground.

b. The angular speed of the pulley can be calculated using the equation Iw2 = (mv2)/2, where I is the rotational inertia, w is the angular speed, m is the mass, and v is the speed. Rearranging the equation, we get w = sqrt((2gh(m2-m1))/(m1+m2+(M/2)))/(MR2)/2. Simplifying, we get w = (1/R) * sqrt((2gh(m2-m1))/(m1+m2+(M/2))). This is the angular speed of the pulley at the moment when Block 2 strikes the ground.

c. The angular displacement