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## Homework Statement

A bead of mass m slides without friction on a rod that is made to rotate at a constant angular velocity ##\omega##. Neglect gravity.

a. Show that ##r=r_0e^{\omega t}## is a possible motion of the bead, where ##r_0## is the initial distance of the bead from the pivot.

b. For the motion described in part a, find the force exerted on the bead by the rod.

c. For the motion described above, find the power exerted by the agency which is turning the rod and show by direct calculation that this power equals the rate of change of kinetic energy of the bead.

## Homework Equations

## The Attempt at a Solution

In the frame fixed to the rotating rod, the force acting on the bead is radially outwards and is of magnitude ##m\omega^2r##. Hence,

$$m\frac{d^2r}{dt^2}=m\omega^2r$$

The solution to the above equation is:

$$r(t)=Ae^{\omega t}+Be^{-\omega t}$$

The motion presented in the question is possible when ##r(0)=r_0## and ##r'(0)=r_0##. Is this correct?

How do I solve the b part? What is the direction of force acting on the bead by the rod?

The bead has both tangential ##(a_t)## and radial ##(a_r)## acceleration.

$$a_t=\frac{dv_t}{dt}$$

where ##v_t## is the tangential velocity. Since ##v_t=\omega r=\omega r_0e^{\omega t}##,

$$a_t=\omega^2r_0e^{\omega t}$$

Also ##a_r=\omega^2r=\omega^2r_0e^{\omega t}##.

Net acceleration: ##\sqrt{a_t^2+a_r^2}##. Hence, force due to rod is ##F=m\sqrt{a_t^2+a_r^2}##. Is this correct?

Any help is appreciated. Thanks!