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# The small block of mass m slides along the radial slot of the disk while the disk rotates in the horizontal plane about its center O. The block is released from rest $\dot{r}$relative to the disk and moves outward with an increasing velocity for the torque M that must be applied to the disk to maintain a constant angular velocity $\omega$ of the disk.

Solution

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We can solve this problem by using the definition of the torque $M$ and it's relation to the angular momentum around $O$, $H_O$. The general expression for the angular momentum around $O$ is:

$H_O = I_O + mr^2 \omega$

Here, $I_O$ is the system's moment of inertia around $O$. The torque $M$ is defined as the first derivative of the angular momentum in respect to time:

\begin{align*} & \dfrac{dH_O}{dt} = \dfrac{d}{dt}(I_O + mr^2 \omega) \\ & M = \dfrac{dI_O}{dt} + \dfrac{d(mr^2 \omega)}{dt} \\ & M = 0 + 2r \dfrac{dr}{dt} m \omega \\ & \boxed{M = 2mr \dot{r} \omega} \end{align*}

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