Moment of inertia of wheel and frictional torque.

  • Thread starter DragonZero
  • Start date
  • #1
12
0

Homework Statement



A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +43 N·m is applied to the wheel for 20 s, giving the wheel an angular velocity of +610 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.

Homework Equations



Rotational Momentum
(0.5)(moment of inertia)(omega^2) = Torque * change in time

The Attempt at a Solution



A) The angular velocity is given in revolutions, so I multiply it by 2 pi in order to get it in radians/sec. I plug that result into the omega of the equation, plugged 43 into torque, and 20 into the time. I know there's more I need to do from there since solving for inertia at this point gave me a result less than 1 which is incorrect, but I'm not sure what that next step is.

B) I think I need to solve for A before I can solve for this.

Thanks. I'm also new to these forums.
 

Answers and Replies

  • #2
Doc Al
Mentor
44,994
1,269
Rotational Momentum
(0.5)(moment of inertia)(omega^2) = Torque * change in time
Careful: The left side is rotational kinetic energy, not momentum.

A) The angular velocity is given in revolutions, so I multiply it by 2 pi in order to get it in radians/sec.
Careful: The angular velocity is given in revolutions per minute.

I plug that result into the omega of the equation, plugged 43 into torque, and 20 into the time. I know there's more I need to do from there since solving for inertia at this point gave me a result less than 1 which is incorrect, but I'm not sure what that next step is.

B) I think I need to solve for A before I can solve for this.
Hint: Instead of trying to solve A and B independently, solve them together. Apply Newton's 2nd law for rotation to the wheel while it's speeding up, and again while it's slowing down. You'll get two equations (and two unknowns) which you will solve together.
 
  • #3
12
0
Ok, so for the angular velocity I did (610 * 2 pi)/60 for radians/sec to get a speed of 63.879.

And I have to use newton's second law, torque = inertia * angular acceleration.

angular velocity = angular acceleration * time

For the speeding up:

63.879 / 20 = angular acceleration = 3.19

For the slowing down, is this correct? 63.879 / 120 = angular acceleration of 0.358.

Then if I put them together it's I * 3.19 = I * 0.358.

I don't think my angular acceleration for the slowing down is correct though.
 
  • #4
Doc Al
Mentor
44,994
1,269
Ok, so for the angular velocity I did (610 * 2 pi)/60 for radians/sec to get a speed of 63.879.
Good.

And I have to use newton's second law, torque = inertia * angular acceleration.
Right. But realize that it's net torque = I * angular acceleration. (When the wheel is speeding up there are two torques acting.)

angular velocity = angular acceleration * time

For the speeding up:

63.879 / 20 = angular acceleration = 3.19

For the slowing down, is this correct? 63.879 / 120 = angular acceleration of 0.358.
Good.

Then if I put them together it's I * 3.19 = I * 0.358.
:confused:


I don't think my angular acceleration for the slowing down is correct though.
That's fine.

Do this:
(1) For the speeding up phase:
Write an expression for the net torque on the wheel, set it equal to I*alpha1
(2) For the slowing down phase:
Write an expression for the torque on the wheel, set it equal to I*alpha2

(Hint: Call the torque due to friction Tf.)
 
  • #5
318
0
WHat is moment of inertia and how do you calculate it? Why would you do so to calculate?
 
  • #6
12
0
For speeding up:

43 - T_f = I * 3.19

For slowing down:

43 = I * 0.358

So what do I do next? Is it

(I * 3.19) + T_f = I * 0.358
 
  • #7
Doc Al
Mentor
44,994
1,269
For speeding up:

43 - T_f = I * 3.19
Good!

For slowing down:

43 = I * 0.358
Careful. When it's slowing down the 43 N·m torque is removed, leaving only the friction.

So what do I do next?
Fix that second equation.
 
  • #8
12
0
For speeding up:

43 - T_f = I * 3.19

For slowing down:

T_f = I * 0.358

Eqeuation:

43 - I * 0.358 = I * 3.19
 
  • #9
Doc Al
Mentor
44,994
1,269
For speeding up:

43 - T_f = I * 3.19

For slowing down:

T_f = I * 0.358

Eqeuation:

43 - I * 0.358 = I * 3.19
That's good. Now solve that last equation for I. (Then you can use that to solve for T_f.)
 
  • #10
12
0
43 = I * (3.19 + 0.357)
I = 12.11

T_f = 12.11 * 0.358 = 4.33
 
  • #11
12
0
It worked! Thank you very much!
 

Related Threads on Moment of inertia of wheel and frictional torque.

Replies
10
Views
20K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
6
Views
8K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
693
  • Last Post
Replies
3
Views
6K
  • Last Post
Replies
9
Views
41K
  • Last Post
Replies
10
Views
6K
  • Last Post
Replies
1
Views
3K
Top