Mastering Physics: Acceleration of a Pulley

In summary, the homework statement states that a string is wrapped around a cylinder of radius r, and that the cylinder can rotate freely about its axis. Find the magnitude α of the angular acceleration of the cylinder as the block descends.
  • #1
danielhep
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1

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.
upload_2017-4-15_19-12-53.png


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.
 
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  • #2
danielhep said:
-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.
 
  • #3
TomHart said:
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.
 
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  • #4
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.
 
  • #5
qdnwedowlnd said:
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 [itex] mg [/itex] 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 [itex] I\ddot{\theta} = Tr [/itex] (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 [itex] \ddot{\theta} [/itex] to check your answer.

Hope that helps.
 
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Related to Mastering Physics: Acceleration of a Pulley

1. What is the acceleration of a pulley?

The acceleration of a pulley is the rate of change of its velocity. It is typically measured in meters per second squared (m/s^2).

2. How is the acceleration of a pulley calculated?

The acceleration of a pulley can be calculated using the equation a = (v2-v1)/t, where v2 is the final velocity, v1 is the initial velocity, and t is the time taken for the change in velocity.

3. What factors affect the acceleration of a pulley?

The acceleration of a pulley can be affected by the mass of the pulley, the mass of the object being pulled, and the force applied to the pulley.

4. How does the angle of the pulley affect its acceleration?

The angle of the pulley can affect its acceleration by changing the direction of the force applied to the pulley. A larger angle may result in a smaller acceleration, while a smaller angle may result in a larger acceleration.

5. What are some real-world applications of understanding the acceleration of a pulley?

Understanding the acceleration of a pulley is important in many industries, such as manufacturing and transportation. It can also be applied in simple machines like elevators and cranes, as well as in more complex systems like roller coasters and car engines.

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