Finding Kinetic Energy and Velocity of a pulley / block system

[pyranos]

Homework Statement

In the problem, a hanging block, m, is attached to a pulley that is mounted on a horizontal axle with negligible friction. The pulley has a mass of M=75kg and a radius of R=6.0cm. The mass of the hanging block is m=1kg. The block is released from rest at a height above the ground of 50cm without the cord slipping on the pulley. The pulley is a uniform disk.

Homework Equations

A) What is the Kinetic Energy of the system right before the block hits the ground?
B) What is the speed of the block right before it hits the ground?

The Attempt at a Solution

I understand that the potential energy of the system in the beginning will be equal to the kinetic energy of the system just before the block hits the ground, however, I do not know how to find the moment of inertia of the system or the velocity of the block.

I have this equation setup:
PE = KE
mgh = 1/2(m)(v^2) + 1/2(I)(ω^2)

I am not sure where to even start since I don't know how to find the moment of inertia, I, or omega. Anything would help out, even a little nudge in the right direction would be appreciated.

 

[grzz]

Moment of inertia of pulley = moment of inertia of a uniform disc = $\frac{1}{2}$Mr$^{2}$.

 

[pyranos]

Moment of inertia of pulley = moment of inertia of a uniform disc = $\frac{1}{2}$Mr$^{2}$.

Would the hanging mass affect the moment of inertia on the disc?

 

[pyranos]

I understand that the moment of inertia for the disc is 1/2(M)(r)^2, but how do you solve the equation for KE? There are two unknown variables still in the equation, and I am not sure how to solve for velocity or omega.

 

[Nickie]

To find the kinetic energy and velocity of the pulley/block system, we can use the conservation of energy principle. As you correctly stated, the potential energy of the system at the starting point will be equal to the kinetic energy just before the block hits the ground.

To find the moment of inertia of the pulley, we can use the formula I = 1/2MR^2, where M is the mass of the pulley and R is the radius. In this case, the moment of inertia would be 0.028125 kg*m^2.

Next, we can use the equation for rotational kinetic energy, KE = 1/2Iω^2, where ω is the angular velocity of the pulley. Since the pulley is a uniform disk, we can use the formula ω = v/R, where v is the linear velocity of the block. Substituting this into the equation for kinetic energy, we get KE = 1/2I(v/R)^2.

Now, we can set the potential energy equal to the kinetic energy and solve for v. This will give us the velocity of the block just before it hits the ground. Plugging in the values from the problem, we get:

mgh = 1/2(m)(v^2) + 1/2(I)(v/R)^2
(1)(9.8)(0.5) = 1/2(1)(v^2) + 1/2(0.028125)(v/0.06)^2
4.9 = 0.5v^2 + (v^2)/0.432
4.9 = 0.5v^2 + 2.31v^2
v^2 = 4.9/2.81
v = √(4.9/2.81)
v = 1.86 m/s

Therefore, the kinetic energy of the system just before the block hits the ground would be 1/2(1)(1.86^2) = 1.73 J and the speed of the block would be 1.86 m/s.

I hope this helps guide you in the right direction for solving the problem. Remember to always use the appropriate equations and units when solving physics problems.