Torque problem - game show wheel

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Homework Help Overview

The discussion revolves around a torque problem involving a solid wheel on a game show that is initially spinning and comes to rest after a certain rotation. Participants are exploring the physics concepts related to angular motion, torque, and moment of inertia.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the torque using the moment of inertia and angular acceleration but questions the correctness of their answer. Other participants ask for clarification on the calculations and the application of formulas, particularly regarding the moment of inertia for a disk.

Discussion Status

Participants are actively engaging with the problem, offering guidance on the correct moment of inertia formula and discussing the implications of angular acceleration and direction. There is a recognition of the need to clarify certain calculations, but no consensus has been reached on the final answer.

Contextual Notes

There is a mention of the original poster's confusion regarding the incorporation of the wheel's rotation through 3/4 of a turn into their calculations. Additionally, the discussion touches on the signs of torque and angular acceleration, indicating potential assumptions about directionality.

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A solid wheel on a game show is given an initial angular speed of 1.25 rad/s in the counterclockwise direction. It comes to rest after rotating through 3/4 of a turn.

Find the torque exerted on the wheel given that it is a disk of radius 0.73 m and mass 6.4 kg.

I got 0.5654 as the answer, but it's not correct. Anyone able to help?
 
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Can you show how you obtained that answer so we can see where you made your mistake?
 
I used mr^2 = (6.4)(0.73)^2 = 3.41056. Then I plugged that into Wf^2 = Wi^2 + 2 (angular acceleration) (delta theta) => 0 = (1.25)^2 + (angular acceleration) (1.5 pi) => 0.1658. Then I multiplied that by 3.41056 to get 0.5654 as my torque.
 
mr2 is the moment of inertia for a point object about a point a distance r away.

You are not dealing with a point, but with an entire disc. The moment of ineria of a disc about the central axis is a slightly different expression.

Here is a helpful website that shows the moments of inertia of various objects: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#cmi

You can either memorize the formulas or derive them by integrating, depending what your teacher/professor requires you to do.
 
Okay, so I use T = 1/2 mr^2. Then once I get the inertia, do I still multiply it by 0.1658, the angular acceleration I got, in order to get torque?
 
Yes, that looks right.
 
Wait...how do I incorporate the 3/4 of a turn into this problem? Do I translate that to 1.5 pi?
 
Yes, you already encorporated that into your work earlier:
FlipStyle1308 said:
I used mr^2 = (6.4)(0.73)^2 = 3.41056. Then I plugged that into Wf^2 = Wi^2 + 2 (angular acceleration) (delta theta) => 0 = (1.25)^2 + (angular acceleration) (1.5 pi) => 0.1658. Then I multiplied that by 3.41056 to get 0.5654 as my torque.
All of that work is right, except that you used mr2 instead of 1/2mr2

Edit: Your answer for acceleration of .1658 looks right but you probably made a typo when posting the equation that's underlined, since your previous equation of "Wf^2 = Wi^2 + 2 (angular acceleration) (delta theta)" is right.
 
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Will my answer end up negative or positive? Cuz my result for the angular acceleration is negative.
 
  • #10
If you define clockwise as negative and counterclockwise as positive it makes sense that the torque is negative because it is acting in the clockwise direction, causing the wheel to stop. You just have to be consistent as to the signs you use.

You are finding the torque excerted that causes the wheel to stop, right? That's what you solved for. If you are trying to find the torque that the person who spun the wheel applied then I don't think you have enough information.
 
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  • #11
Never mind, I got it. Thank you for your help!
 
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