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Moment of Inertia and torque?

by miamirulz29
Tags: inertia, moment, torque
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miamirulz29
#1
Dec6-09, 04:22 PM
P: 62
1. The problem statement, all variables and given/known data
The combination of an applied force and a frictional force produces a constant total torque of 39.9Nm on a wheel rotating about a fixed axis. The applied force acts for 4.5s, during which time the angular speed of the wheel increases from 3 rad/s to 12 rad/s. The applied force is then removed. The wheel comes to rest in 72s.
A. What is the moment of inertia of the wheel? Answer in units of kgm^2
B. What is the magnitude of the frictional torque? Answer in Nm
C. What is the total number of revolutions of the wheel?


2. Relevant equations
[tex]\sum[/tex][tex]\tau[/tex] = I(moment of inertia) * [tex]\alpha[/tex]



3. The attempt at a solution
A. 39.9 = 2I
I = 19.95
That is correct.
B. T = (19.95)(9/72) = 2.49375
That is incorrect. What am I doing wrong?
C. I need some help, do not know where to start.
Thanks in advance.
Sorry if the equations look bad. This the first time I am using Latex and I still don't know exactly how to use it correctly.
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miamirulz29
#2
Dec6-09, 04:29 PM
P: 62
Well I just I figured out Part b. Instead of doing 12-3/72, I had to do 12/72. Can somebody explain that to me please.
nrqed
#3
Dec6-09, 04:33 PM
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Quote Quote by miamirulz29 View Post
Well I just I figured out Part b. Instead of doing 12-3/72, I had to do 12/72. Can somebody explain that to me please.
You need to consider the part of the motion when only the frictional torque is acting. The wheel then slows down from 12 rad/s to 0 rad/s in 72 sec. This is why you need to divide 12 rad/s by 72 s to get alpha

miamirulz29
#4
Dec6-09, 04:34 PM
P: 62
Moment of Inertia and torque?

Oh right, thank you. Any help for part C? Just a way for me to get started please.
nrqed
#5
Dec6-09, 04:37 PM
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Quote Quote by miamirulz29 View Post
Oh right, thank you. Any help for part C? Just a way for me to get started please.
You're welcome.

First you need to find the total angle of rotation during the acceleration part and during the deceleration part (you must have seen the formula [itex] \theta = \omega_i t + 1/2 \alpha t^2[/itex]). Then add the two angles for the total angle and fivide by 2 Pi to get the number of revolutions
miamirulz29
#6
Dec6-09, 05:38 PM
P: 62
So could I do this: theta = (1/2)(12/72)(72^2) for the decelerating part and for the accelerating part could I use the other formula: 12^2 - 9^2 / 2(12-3/4.5). Then add those together and divide by 2pi?
nrqed
#7
Dec6-09, 06:36 PM
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Quote Quote by miamirulz29 View Post
So could I do this: theta = (1/2)(12/72)(72^2) for the decelerating part and for the accelerating part could I use the other formula: 12^2 - 9^2 / 2(12-3/4.5). Then add those together and divide by 2pi?
For the decelerating part, you are missing one term since omega_i is not zero.

For the accelerating part, it sounds good except that you mant 3^2 instead of 9^2.
miamirulz29
#8
Dec6-09, 06:41 PM
P: 62
Yes I meant 3^2 instead of 9^2. But isn't omega_i zero because is it come to rest when it decelerates.
nrqed
#9
Dec6-09, 07:04 PM
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Quote Quote by miamirulz29 View Post
Yes I meant 3^2 instead of 9^2. But isn't omega_i zero because is it come to rest when it decelerates.
It comes to rest at the end of the decelarating part so omega final is zero. But when it started decelarating, its omega was 12 rad/s, so that's the value of omega_i for the decelerating pat


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