Question

Consider an eraser, weighing 20g, sitting 5cm from the center of a turntable (flat disk attached to a motor which can spin the disk up or down). If the coefficient of static friction between the eraser and the turntable is 0.1, and the turntable is slowly spun-up just shy of the spin rate at which the eraser starts to slip
A.Calculate how many revolutions the turntable will make every minute at the end (when the eraser is just about to slip). Show your reasoning!
B.What force is responsible for making the eraser go faster while the turntable is spinning up?

Answer 1

Answer:

A. 42.3 revolution per minute

B. Centripetal Force

Explanation:

20 g = 0.02 kg

5 cm = 0.05 m

Let g = 9.8 m/s2

A.The friction force acting on the eraser is the product of normal force and its coefficient:

$F_f = N\mu = mg\mu = 0.02*9.8*0.1 = 0.0196 N$

For the eraser to begin slipping, its centripetal acceleration must win over the acceleration created by friction force, which is

$a_f = F_f / m = 0.0196 / 0.02 = 0.98 m/s^2 = a_c$

We can calculate the angular velocity from this:

$a_c = \omega^2 r$

where r = 0.05 is the radius of rotation, which is distance from the center to the eraser

$\omega^2 = a_c/r = 0.98 / 0.05 = 19.6$

$\omega = \sqrt{19.6} = 4.43 rad/s$

If it spins and angle of 4.43 rad per second then for every minute (60 seconds) it would spin an angle of 4.43 * 60 = 265.6 rad/min

Since 1 revolution is 2π rad, then in a minute it would spin 265.6 / 2π = 42.3 revolution/minute

B. The force responsible for making the eraser go faster while the turntable is spinning up would be the centripetal force.