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Question

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 r˙\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 ω\omega of the disk.

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Answered 2 years ago
Answered 2 years ago
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We can solve this problem by using the definition of the torque MM and it's relation to the angular momentum around OO, HOH_O. The general expression for the angular momentum around OO is:

HO=IO+mr2ωH_O = I_O + mr^2 \omega

Here, IOI_O is the system's moment of inertia around OO. The torque MM is defined as the first derivative of the angular momentum in respect to time:

dHOdt=ddt(IO+mr2ω)M=dIOdt+d(mr2ω)dtM=0+2rdrdtmωM=2mrr˙ω\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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