Register to reply 
Moment of inertia in a wheel? 
Share this thread: 
#1
Mar2112, 10:12 PM

P: 3

1. The problem statement, all variables and given/known data
So basically there is a wheel with a radius of 0.3 m. A light cord wrapped around the wheel supports a 0.75kg object that accelerates at 3 m/s^2 downwards. What is the moment of inertia of the wheel? (no friction) 2. Relevant equations I=mr^2 T=I(alpha) (alpha)=ra 3. The attempt at a solution Seems like it should be an easy question, but the inclusion of the acceleration threw me off. The two approaches I have thought about are: I=mr^2, I=(0.3*0.75^2) which gives an option in the answers (0.15), but that seems too easy and I did not use the acceleration. The other approach is using Torque=I (angular acceleration), using the above value for inertia (0.15), and then calculating angular acceleration by r * linear acceleration (0.3*3), but this produced a number not in the solution list (0.135). Are there other approaches I have missed? Involving forces or the like? 


#2
Mar2112, 10:42 PM

Emeritus
Sci Advisor
PF Gold
P: 5,197

You know the torque around the centre of the wheel that is produced by the weight of the hanging mass. You also know the angular acceleration of the wheel because, as you pointed out, it is related to the linear acceleration of the mass. So, given both torque and angular acceleration, you can solve for the moment of inertia. EDIT: except it you should have written: angular acceleration = (linear acceleration)/r divided by r, not multiplied by r. 


#3
Mar2112, 11:01 PM

P: 3

Ohhhh right. Wow, quite a fail on my behalf.
So now I calculated Torque as F*d ((3*0.75)*0.3) and got 0.675. Then I divided this by 10 (calculated angular acceleration  3/0.3) and got 0.068. This is an option...but is it correct? Or should I factor in gravity to my force calculation? 


#4
Mar2112, 11:05 PM

Emeritus
Sci Advisor
PF Gold
P: 5,197

Moment of inertia in a wheel?
Not only are units am essential part of any meaningful physical calculation, but they also help you catch errors. If the units don't work out, you've made a mistake. Your final answer should be in Nm, since you're computing a torque. 


#5
Mar2112, 11:17 PM

Emeritus
Sci Advisor
PF Gold
P: 5,197

On second thought...since the mass is not in free fall, the net force on it must be smaller than its weight. (The net force is ma, and a is smaller than the acceleration due to gravity) This means that the rope must be pulling up on the mass to counter its weight partly.
So think maybe the tangential force that is producing the torque is equal to the net force after all, and not to the weight. Hmm...I'll have to think about it some more. 


#6
Mar2212, 08:57 AM

Mentor
P: 11,625

Hint: The 0.75kg object is not falling at g, so there must be a tension T in the cord that is slowing its fall. This same tension will be responsible for the torque on the wheel.



#7
Mar2212, 10:15 AM

Emeritus
Sci Advisor
PF Gold
P: 5,197

So I guess the OP has to solve for the tension in the rope by figuring out the difference between the downward force on the mass due to gravity, and the net force that ends up acting on it. 


#8
Mar2212, 10:21 AM

Emeritus
Sci Advisor
PF Gold
P: 5,197

Actually, hang on. What if the mass WERE in freefall? Then your analysis would lead us to believe the tension in the rope would be zero. Where, then, would the tangential force that is spinning up the wheel be coming from?



#9
Mar2212, 10:33 AM

Mentor
P: 11,625




Register to reply 
Related Discussions  
Moment of inertia wheel and bug  Introductory Physics Homework  5  
Determine the moment of inertia of the wheel  Introductory Physics Homework  2  
Moment of inertia for wheel  General Physics  6  
Moment of Inertia for a wheel  Introductory Physics Homework  3  
Moment of Inertia of a Wagon Wheel  Introductory Physics Homework  2 