Atwood's Machine: Calculating Velocity and Angular Velocity

[icf927]

Homework Statement

A horizontal cylinder on frictionless bearings has a moment of inertia of 0.8kg*m^2 and a radius of 22cm. A 15 kg mass is attached to a 8kg mass with a massless string wrapped around the cylinder. The string does not slip on the cylinder. If the 15 kg mass is released from rest 3.4m above the floor,
a) what is the velocity of the 8kg mass when the 15kg mass hits the floor?
b) what is the angular velocity of the cylinder when the 15kg mass hits the floor?

Homework Equations

Mg-T=Ma
T-mg=ma
V^2=V0^2 + 2a(y-y0)
mgh=1/2mv^2 + 1/2Iω^2

The Attempt at a Solution

a)I solved for T in one of the above equations and plugged into the other equation. Then I plugged in the masses and solved for the acceleration which should be the same for both boxes and I got 2.9826 m/s^2 for a. I then used the third equation given that the original velocity is zero, the height is 3.4m, and the acceleration calculated earlier to be 2.9826, and calculated the final velocity which I got as 4.50m/s.
b) I used conservation of energy and used the fourth equation. I plugged rω in for v and then plugged in for the variables known and solved for ω which I got to be 25.6 rad/s.

 

[ehild]

Homework Statement

A horizontal cylinder on frictionless bearings has a moment of inertia of 0.8kg*m^2 and a radius of 22cm. A 15 kg mass is attached to a 8kg mass with a massless string wrapped around the cylinder. The string does not slip on the cylinder. If the 15 kg mass is released from rest 3.4m above the floor,
a) what is the velocity of the 8kg mass when the 15kg mass hits the floor?
b) what is the angular velocity of the cylinder when the 15kg mass hits the floor?

Homework Equations

Mg-T=Ma
T-mg=ma
V^2=V0^2 + 2a(y-y0)
mgh=1/2mv^2 + 1/2Iω^2

The Attempt at a Solution

a)I solved for T in one of the above equations and plugged into the other equation. Then I plugged in the masses and solved for the acceleration which should be the same for both boxes and I got 2.9826 m/s^2 for a. I then used the third equation given that the original velocity is zero, the height is 3.4m, and the acceleration calculated earlier to be 2.9826, and calculated the final velocity which I got as 4.50m/s.
b) I used conservation of energy and used the fourth equation. I plugged rω in for v and then plugged in for the variables known and solved for ω which I got to be 25.6 rad/s.

The cylinder has mass and moment of inertia. To accelerate it, torque is needed: The tensions are not the same at both sides.

Your last equation involves only one mass, but there are two of them, moving in opposite directions.

You need to use only conservation of energy and the "no slip" condition.

ehild

 

[icf927]

How do I do that?

 

[ehild]

How do I do that?

You know what conservation of energy means?
No slip means that the speed of the rope (and that of the masses) is v= rω.

ehild

 

[icf927]

yes but I do not know how to use that to find the answers.

 

[ehild]

Let be the potential energy zero at the initial height of the blocks. The heavy one drops 3.4 m; the light one rises 3.4 m. What is the new potential energy?

I hope you can write up the total kinetic energy of the two blocks and the rotational energy of the cylinder.

ehild

 

[icf927]

I wrote up the equation and got Mgh=1/2mv^2+1/2Mv^2+1/2Iw^2+mgh
where M=15kg and m=8kg. is this right?

 

[ehild]

It is right. Replace w=v/r, and solve for v.

ehild

 

[icf927]

i got a velocity of 3.44m/s

 

[icf927]

how do you go about solving part b though?

 

[ehild]

w=v/r

ehild

 

[icf927]

I got 3.44 m/s for part a and 15.6 rad/s for part b. is that what you got?

 

[ehild]

I got 3.44 m/s for part a and 15.6 rad/s for part b. is that what you got?

Yes

ehild