Mastering Physics: Acceleration of a Pulley

[danielhep]

Homework Statement

A string is wrapped around a uniform solid cylinder of radius r, as shown in (Figure 1) . The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass m.
Find the magnitude α of the angular acceleration of the cylinder as the block descends.

Homework Equations

F=ma=mg-T
Torque=Iα=-Tr
a=-rα
I=mr²/2

The Attempt at a Solution

I use these three equations to eliminate T and solve for α.

m(-rα)=mg-T
-1/2mr=T
-mrα=mg+1/2mr
-α=g/r+1/2
α=-1/2-g/r

I tried this answer however it was wrong.

Mastering says the correct answer is
α = 2g/3r

I'm really not sure where this comes from, so any help would be great! I need to understand where I went wrong.

 

[TomHart]

-1/2mr=T

Where did this come from? It doesn't look right.

Edit: Okay, now I see what you did. You were equating Iα to -Tr, then substituting I = mr²/2 for I. You should work through that algebra again and I think you'll find your problem.

 

[danielhep]

Where did this come from? It doesn't look right.

Edit: Okay, now I see what you did. You were equating Iα to -Tr, then substituting I = mr²/2 for I. You should work through that algebra again and I think you'll find your problem.

I dropped the g. Thank you so much! I found the correct answer.

 

[qdnwedowlnd]

why is Iα = -Tr. Shouldn't the force acting on it be mg since T (from cylinder to block) - T (from block to cylinder) -mg = mg? Or shouldn't it be mg - T, which is what we defined the net force as above.

 

[Master1022]

why is Iα = -Tr. Shouldn't the force acting on it be mg since T (from cylinder to block) - T (from block to cylinder) -mg = mg? Or shouldn't it be mg - T, which is what we defined the net force as above.

It won't just be mg as that would assume that there is no acceleration of the mass.

The pulley experiences a torque just due to the Tension, so that is why I\ddot{\theta} = Tr (or with a minus sign if you choose to define in other direction), and then the other terms come into play when we substitute our expression that we get from Newton's 2nd law on the mass.

Note: you can also solve this by energy and differentiate to get \ddot{\theta} to check your answer.

Hope that helps.