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Find Moment of Inertia and Frictional Torque

  • Thread starter personguy
  • Start date
6
0
1. A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +52 N m is applied to the wheel for 22 s, giving the wheel an angular velocity of +530 rev/min. The external torque is then removed, and the wheel comes to rest 120 s later. (Include the sign in your answers.)


(a) Find the moment of inertia of the wheel.

(b) Find the frictional torque, which is assumed to be constant.

I have no idea where to begin with this... I need a mass and radius to find the moment of inertia.....but they aren't given.
 
Last edited:
1,356
0
Yes, the mass and the radius of the wheel are not given so you're going to have to derive them from the information given. Note that you're given torque, time and angular velocity. Do you know of any equations involving the latter that you can use to derive the mass and the radius?
 
6
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No, I don't.
 
6
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I did, I don't see anything.
 
1,356
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Looks like I'm going to have to be a bit more specific: In your book, what are the equations for torque and angular velocity?
 
6
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Angular Velocity: d theta / dt

Torque: Tangential force times radial distance (Ft times r) = Fr sin theta = Fl
 
6
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My friend tried it this way, but it didn't give the right answers:

Torque = F.d = I.a

I = moment of inertia

a = angular acceleration

F = force

d = distance.

The first torque made the wheel rotate from the rest until the speed gained an angular velocity of 530 rev / min

The torque is 52Nm

52 Nm = I.a

I = moment of inertia of the wheel

a = angular aceleration

let's find a, with : Wf = 530 rev / min, time = 22s, initial velocity = 0 m/s

530*2pi / 60 = a*22

a = 2.52 rad/s^2

then : 52 = I*2.52 >>> I = 20.63 kg*m^2

b), Let's find the frictional torque with :

Torque = 20.63.a'

a' = final angular acceleration ( when the wheel stops)

initial speed before the braking : 530*2pi / 60 rad / s

final velocity = 0 rad / s, time = 120 s

0 = 530*2pi/60 - a'120

a' = 0.46 rad / s^2

Torque = 20.63*0.46 = 9.54 Nm
 
1,356
0
The first torque made the wheel rotate from the rest until the speed gained an angular velocity of 530 rev / min

The torque is 52Nm
Note that the net torque on the wheel is the sum of the external torque and the frictional torque. You seem to be ignoring the frictional torque when finding the angular acceleration of the wheel.
 
6
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So where would that come into this problem?
 
1,356
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So where would that come into this problem?
That's from the initial analysis of the problem. Do you agree with my statement? Do you see how the net torque on the wheel for the first 22 seconds equals the external torque plus the frictional torque?
 

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