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A roller of radius 10cm is spinning with a angular velocity of 15 rad/s. It has a completely inelastic collision with a hunk of clay, with mass m moving at 3m/s at it's very bottom edge. Does the angular velocity of the roller (now with stuck clay) increase, decrease or stay the same? (The picture should clarify)

I think it increases because at the moment of the collision there is a torque on the roller, but the moment of inertia also increase so I am not sure. Thanks in advance for any help!

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    $\begingroup$ You can conserve angular momentum about center of roller and check. $\endgroup$
    – SarGe
    Jun 1, 2020 at 2:45
  • $\begingroup$ media.discordapp.net/attachments/689905951104565312/… I set up an equation but it's not clear what the change in W is $\endgroup$
    – Mustafa
    Jun 1, 2020 at 3:31
  • $\begingroup$ i think you need to consider the torque produced once it hit,if you can take the momentum along the centre I think due to addition of an extra mass there will be a change in velocity of the rotating body $\endgroup$ Jun 1, 2020 at 3:37

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Angular momentum is conserved in an inelastic collision, so all you have to do is calculate the total angular momentum of the system the moment before the collision and the moment after. Before the collision, the total angular momentum is the sum of the angular momentum of the roller and the angular momentum of the clay (as measured from the axis of the roller) right before it hits the roller. After the collision, you can once again calculate the total angular momentum of the system. The new moment of inertia is different, but it is easy to calculate. All you have to do is add the moment of inertia due to the extra spinning mass of clay. Set the final angular momentum equal to the angular momentum before the collision, and you have your answer.

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Using the dynamical equation for the angular momentum of a system, \begin{equation} \dfrac{d \mathbf{\Gamma}_H}{dt} = - \dot{\mathbf{x}}_H \times \mathbf{Q} + \mathbf{M}_H^{ext} \ , \end{equation} neglecting gravity, and choosing the pole $H$ as the center of rotation of the disk (in order to have $\mathbf{M}_H = \mathbf{0}$), you get the conservation of the angular momentum of the system, \begin{equation} \dfrac{d \mathbf{\Gamma}_H}{dt} = \mathbf{0} \ . \end{equation} Now, evaluating the angular momentum before and after the collision, considering only the component of the angular momentum perpendicular to the plane of motion, we get \begin{equation} \Gamma_{H,0} = I_d \Omega_0 + m R v_0 = \left[ I_d + \alpha m R^2 \right] \Omega_0 \\ \Gamma_{H,1} = I_d \Omega_1 + m R v_1 = \left[ I_d + m R^2 \right] \Omega_1 \ , \end{equation} having used the kinematic constraint $v_1 = R \Omega_1$ after the collision, and the definition $\frac{v_0}{R \Omega_0} = \alpha$, with the coefficient $\alpha$ using to explicitly compare the velocity $v_0$ with the tangential velocity of the disk $R \Omega_0$ ($\alpha > 1$ means point mass faster than the tangential velocity).

Equating the two expressions of the angular momentum before and after the collision, we get \begin{equation} \dfrac{\Omega_1}{\Omega_0} = \dfrac{I_d + \alpha m R^2}{I_d + m R^2} \ , \end{equation} from which it's easy to deduce that $\Omega_1 > \Omega_0$ if $\alpha > 1$, and viceversa.

In the problem, \begin{equation} \alpha = \dfrac{v_0}{R \Omega_0} = \dfrac{3.0 \, m/s}{0.1 \, m \cdot 15 \, rad/s} = \dfrac{3.0 \, m/s}{1.5 \, m/s} = 2 \ . \end{equation}

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By using conservation of momentum the angular velocity increases because the clay apply a torque about centre which moves the spinning disk fastly

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  • $\begingroup$ This might not be always true because the total moment of inertia also increases. $\endgroup$ Jun 1, 2020 at 4:30
  • $\begingroup$ @VincentThacker Yet the clay is slowed down somewhat, so it must have transferred some momentum to the disk. $\endgroup$
    – Zac67
    Mar 18, 2022 at 12:22

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