Calculating Rotational Work on a Solid Cylinder

In summary, the tension is the same for both the pulley and the mass, so it can be used in both equations to solve for the acceleration.
  • #1
Dethrone
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0

Homework Statement



Calculate the work done on a solid cylinder of mass 5.0 kg and radius 55.6 cm by an applied force of 12.2N if it turns 45 degrees.

Homework Equations



W = torque x angle (radians)

The Attempt at a Solution



I just used:

W = torque x angle
= Force x perpendicular radius x angle
= 12.2 x 55.6 x pi/4

Why is this wrong? I think it's wrong, because part b of the question asks how the answer would change if the object was a ring.
 
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  • #2
Why do you think it would matter if a ring were used?
 
  • #3
Because they would have different moment of inertias. But, I didn't even need to use it for the problem, so could the answer be that it would be the same regardless of the object shape?
 
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  • #4
I would think so.
 
  • #5
Thanks! i have a second question:



At 23:50 of the video, he says that torque = tension (rope) x radius. Why is that? Wouldn't the torque = force of gravity x radius? Since the force of gravity is the only force acting downwards and the tension forces cancel?
 
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  • #6
If the pulley was fixed and did not move, then the tension T would indeed be equal to mg. However, if the pulley is released, then the tension would be less than mg because the pulley is accelerating. Think what the tension in the rope would be if the pulley had no moment of inertia or very little mass.
 
  • #7
If the rope had very little mass, I would guess that the tension of the rope would be equal to the force of gravity? But I'm still confused, doesn't the force of gravity have any effect on the torque? since it's contributes to the force that the pulley experiences

In this old thread: https://www.physicsforums.com/showthread.php?t=319573
The guy did the force of gravity x radius
 
  • #8
Yes, if there were no gravity then there would be no tension. As I said, if the pulley was locked and could not move then the tension would be mg. But if the pulley were allowed to move, then the tension would be less than mg.
 
  • #9
I'm just confused with your last statement. "But if the pulley were allowed to move, then the tension would be less than mg." I understand that tension would be less than mg, but isn't mg ultimately the force pulling on the pulley since it's attached to the rope also? So wouldn't it be that force (mg) x radius? Or is it that while mg is the force that's keeping the pulley moving, the force that the pulley ACTUALLY experiences is the tension force?

I just had another idea: Is it that tension x radius equals the torque around the rotating disc and force of gravity x radius is the torque around the whole system?
 
  • #10
Not sure about your last statement but consider the following. Draw a free body diagram of the mass hanging on the rope. If the pulley was locked (not allowed to rotate) then the mass would not move and the tension T woujld equal mg, T balances mg. Now, unlock the pully. The mass will start to accelerate downward and the pulley will start to rotate. The ONLY way the mass could start accelerating downward is if mg is greater than T the tension.
 
  • #11
Okay, I understand both now, but is the reason why we multiply the tension by radius because the force that the pulley actually experiences is the tension? Because it's my understanding that Torque = Force x radius, so when I first attempt this problem, I thought that mg was the actually force that's moving the pulley.
 
  • #12
Dethrone said:
= 12.2 x 55.6 x pi/4

Why is this wrong?
What units do you think your answer is in? Check the units of all the given data.
 
  • #13
I know, 55.6 should be in m. My answer should be in N m.

I still need this question resolved, though,

Dethrone said:
Okay, I understand both now, but is the reason why we multiply the tension by radius because the force that the pulley actually experiences is the tension? Because it's my understanding that Torque = Force x radius, so when I first attempt this problem, I thought that mg was the actually force that's moving the pulley.
 
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  • #14
I am confused as to what question you are asking about, part a or part b? For part a the torque = tension (or force) X radius and the answer will be in n-m. What is your question about part b?
 
  • #15
is the reason why we multiply the tension by radius because the force that the pulley actually experiences is the tension? What I don't get is if the force that's actually moving the pulley is the force of gravity, WHY do we multiple the tension force by radius and not mg by radius?
 
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  • #16
That is correct. The Tension, T, acts downward on the pulley and upwards on the mass, m, but the tension is the same in both cases. To solve this problem (shown in the video at 23:50, you create two situations. One relating the tension to the acceleration of the pulley, and the second relating the tension to the acceleration of the hanging mass. Then if you set the tensions equal to each other, you can find the acceleration of the lower mass.
 

FAQ: Calculating Rotational Work on a Solid Cylinder

What is rotational work?

Rotational work is the amount of energy transferred to an object as a result of a rotational force acting on it.

How is rotational work calculated?

Rotational work is calculated by multiplying the force applied to an object by the distance traveled in the direction of the force, and the angle between the force and the displacement.

What is the unit of measurement for rotational work?

The unit of measurement for rotational work is the joule (J), which is equivalent to a newton-meter (N*m).

What is the difference between rotational work and linear work?

Rotational work involves an object rotating around an axis, while linear work involves an object moving in a straight line. The calculation for rotational work takes into account the angle of rotation, while linear work does not.

What are some real-world applications of calculating rotational work?

Calculating rotational work is useful in understanding the efficiency of machines such as motors, turbines, and propellers. It is also important in sports such as gymnastics and figure skating, where rotational motion is involved.

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