Rotational Energy/Speed of a System?

[Juniper7]

Homework Statement

A solid sphere of mass 6.0 kg is mounted on a vertical axis and can rotate freely without friction. A massless cord is wrapped around the middle of the sphere and passes over a 1.0 kg pulley and is attached to a block of mass 4.0kg. What is the speed of the block after it has fallen 80 cm? Treat the pulley as a sold cylinder.

Homework Equations

I of a solid cylinder = 1/2MR²
I of a sphere = 2/5MR²
KE_rotational = 1/2Iω²
KE_linear = 1/2mv²
PE = mgh
ω = v/r

The Attempt at a Solution

I am very lost in this question. I have tried to calculate the kinetic energies of each part of the system but it doesn't give radii or ω so I get stuck. I thought this could be relevant:

KE_sphere + KE_pulley + KE_block

But what would I equate those energies to? Thank you in advance for any help!

 

[Juniper7]

I think I may have worked through it but it doesn't seem right:

KE_sphere = 1/2Iω² = (1/2)(2/5mr²)ω² = (1/5)(6kg)r²ω²
then substitue in ω = v/r and get KE_sphere =(1.2)v_sphere²

I did this for each the pulley and the block and got:

KE_pulley = (1/4)v_pulley²
KE_block = 2v_block²

Then I thought I could use this equation:

PE_block = KE_sphere + KE_pulley + KE_block

Because the sphere and pulley don't have potential energy, right?

m_blockgh = (1.2)v_sphere² + (1/4)v_pulley² + 2v_block²

(4kg)(9.81m/s²)(0.8m) = 1.2v² + 0.25v² + 2v²
v = 3.0m/s

The part I'm really not sure about is the velocities of each part of the system. Will they all be the same? I assumed they are all the same to get my answer

 

[SammyS]

Homework Statement

A solid sphere of mass 6.0 kg is mounted on a vertical axis and can rotate freely without friction. A massless cord is wrapped around the middle of the sphere and passes over a 1.0 kg pulley and is attached to a block of mass 4.0kg. What is the speed of the block after it has fallen 80 cm? Treat the pulley as a sold cylinder.

Homework Equations

I of a solid cylinder = 1/2MR²
I of a sphere = 2/5MR²
KE_rotational = 1/2Iω²
KE_linear = 1/2mv²
PE = mgh
ω = v/r

The Attempt at a Solution

I am very lost in this question. I have tried to calculate the kinetic energies of each part of the system but it doesn't give radii or ω so I get stuck. I thought this could be relevant:

KE_sphere + KE_pulley + KE_block

But what would I equate those energies to? Thank you in advance for any help!

Of course, those are two different radii, and two different moments of inertia. -- Use subscripts or something similar.

The image is below.

I'll answer your 2nd post presently.

 

[SammyS]

I think I may have worked through it but it doesn't seem right:

KE_sphere = 1/2Iω² = (1/2)(2/5mr²)ω² = (1/5)(6kg)r²ω²
then substitue in ω = v/r and get KE_sphere =(1.2)v_sphere²

I did this for each the pulley and the block and got:

KE_pulley = (1/4)v_pulley²
KE_block = 2v_block²

Then I thought I could use this equation:

PE_block = KE_sphere + KE_pulley + KE_block

Because the sphere and pulley don't have potential energy, right?

m_blockgh = (1.2)v_sphere² + (1/4)v_pulley² + 2v_block²

(4kg)(9.81m/s²)(0.8m) = 1.2v² + 0.25v² + 2v²
v = 3.0m/s

The part I'm really not sure about is the velocities of each part of the system. Will they all be the same? I assumed they are all the same to get my answer

The sphere and pulley are stationary; they don't have velocity.

The only objects in this problem with linear velocity are the block and cord. The cord presumably doesn't slip on the surfaces of the pulley or the sphere.

What do you think the velocities you used are relevant to?

 

[Juniper7]

The sphere and pulley are stationary; they don't have velocity.

The only objects in this problem with linear velocity are the block and cord. The cord presumably doesn't slip on the surfaces of the pulley or the sphere.

What do you think the velocities you used are relevant to?

Are the velocities relevant to the speed the sphere turns at and the speed the pulley turns at?

 

[SammyS]

Are the velocities relevant to the speed the sphere turns at and the speed the pulley turns at?

Those velocities are the velocities of the surfaces of the speed & pulley at the points they contact the cord. Thus they're the velocity of the cord, which happens to be fastened to the block. Right >\?

 

[Juniper7]

Oh ok. So because the cord is attached to the block, they should both have the same velocities?

 

[SammyS]

Oh ok. So because the cord is attached to the block, they should both have the same velocities?

Yes.

 

[Juniper7]

Thank you!