Rotational Inertia Problem

1. Oct 28, 2013

xtrubambinoxpr

1. The problem statement, all variables and given/known data

in Attachment

2. Relevant equations

I = .5*mr^2

3. The attempt at a solution

I have no idea what is going on. This is a pre section quiz and I guessed on one and got it right. I dont know what is meant by the equations not giving I. I googled and noticed I has various equations. I plugged in the one I though most pertinent but it was marked wrong. Someone assist me please!

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2. Oct 28, 2013

johntcmb

The moment of inertia, I should be .5*MR^2, not .5*mR^2. See if that helps.

3. Oct 28, 2013

xtrubambinoxpr

I don't see how. I tried calculating with both m and M but I was wrong both times

4. Oct 28, 2013

haruspex

The wheel is not stated to be a uniform disc. That is why I is given as an input variable. It cannot be obtained from a formula. But the question setter seems to have forgotten to provide the numerical value of I.

5. Oct 28, 2013

xtrubambinoxpr

Are you sure? Lol
Not to insult you but googling it has the same question solved with I not given. And I can figure it out for the life of me :/

6. Oct 28, 2013

johntcmb

Since you "guessed" the answer for a correctly, you can substitute that value into eq. 3 and with a little alfebra get T = mg - ma. Using that, you get T = 4.524 N. Plugging I=.5*MR^2 into eq. 4 should give you the same result.

7. Oct 28, 2013

nil1996

I used the equations

mg-T=ma

TR=Iα

a=Rα

Solve these equations you will get the answer

8. Oct 28, 2013

haruspex

There are 6 variables there: m, T, R, a, I, α. You have three equations. Therefore you need the values of three variables. You are told the numerical values of m and R, that's all. You must have used a fourth equation.

9. Oct 28, 2013

haruspex

Quite sure. Note that equation (4) has I in the final formula before evaluation, implying you are supposed to substitute a number for I. If you were expected to obtain I from M and R there would be an equation for that somewhere in the text.
If you think you have found somewhere on the net that solves this on the information given, pls post the link.

10. Oct 28, 2013

nil1996

we also have value of I
I have assumed it to be a thin disc.

Last edited: Oct 28, 2013
11. Oct 28, 2013

haruspex

You are not given that information, nor a formula. Instead, I is given symbolically as an input and also appears in the final equations. This strongly suggests it is not supposed to be a uniform disc, but the numerical value was to be given.
If you read back through the posts you'll see that xtrubambinoxpr tried using your assumption (though once with the wrong mass) and the answer was rejected.

12. Oct 29, 2013

nil1996

But the acceleration looks right

13. Oct 29, 2013

haruspex

In what way does it 'look right'? I don't see the answer supplied anywhere in the thread.

14. Oct 29, 2013

nil1996

See in the attachment he has got acceleration right

15. Oct 29, 2013

haruspex

Ok, I see. It could be rounding error, i.e. 4 is accepted as close enough.
xtrubambinoxpr, it accepted a=4. Since α = a/R, if a is right then it should be easy to get α right. Similarly, T = mg-ma. What did you enter for α and T?