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This Question is from my practice Midterm.

A wheel with Radius R, mass m, moment of Inertia I = $1/3 mR^2$ A ball of Clay of mass m, and initial Velocity v

The wheel has a cup on it which catches the Clay ball. Solve for the final angular velocity of the wheel. we are told to neglect the Cup attached to the wheel.

My Attempt:

momentum of the ball = $mv$.

angular momentum of the wheel = $I\omega$

since all the energy from the Ball goes to the Wheel, I equate the two.

$$mv=(I\omega)$$ where now, because of mass= 2m, I = $2/3mR^2$

$$mv=\frac {2}{3}mR^2\omega$$ $$\omega = \frac {3mv}{2mR^2}= \frac {3v}{2R^2}$$

But the Correct Answer is $$\frac {3v}{4R}$$

Please help!

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  • $\begingroup$ Hi! You are equating the angular momentum of the wheel to the linear momentum of the ball. This is clearly wrong, and as a consequence your answer does not even have the correct units. $\endgroup$ – DelCrosB Oct 29 '16 at 23:21
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If I'm picturing this correctly, the wheel will begin to rotate once it catches the ball. If this is the case, you need to work with angular momentum, from the moment the ball is caught.

The initial angular momentum will be $L = mvR$

Secondly, the moment of inertia should be a sum of the moments of the wheel and the ball.

$I_{wheel} = \frac{1}{3}mR^2$

$I_{ball} = mR^2$ (if modeled as a point mass)

Then, $I_{total} = \frac{1}{3}mR^2+mR^2 = \frac{4}{3}mR^2$

So from here, this should give you the correct answer:

$mvR = \omega I_{total} = \frac{4}{3}mR^2\omega $

$v = \frac{4}{3}R\omega $

$$\omega = \frac{3v}{4R}$$

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  • $\begingroup$ I didn't know that the mass of the ball could be written as a moment of Inertia. Thanks for your help! $\endgroup$ – Jess L Oct 29 '16 at 23:25
  • $\begingroup$ Also, Why did you define angular momentum as mvR? When I google formula for Angular Velocity, I get $\omega = v/R$ so shouldn't Angular momentum = $mv/R?$ $\endgroup$ – Jess L Oct 29 '16 at 23:28
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    $\begingroup$ Well, the definition for the moment of inertia for a point mass is $I = r^2 m$. As $\omega = \frac{v}{r}$. So don't forget that angular momentum is defined as $L = I\omega$, so if we substitute these definitions into the equation for $L$, we obtain $L = (mr^2)(\frac{v}{r})$ which gives us $L = mvr$ after cancelling. $\endgroup$ – bleuofblue Oct 29 '16 at 23:34

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