Question

A horizontal circular turntable rotates about its center at the uniform rate of 10.9 revolutions per minutes. Find the greatest distance (in meter) from the center at which a small block will remain stationary relative to the turn table.The coefficient of static friction between the turntable and the object is 0.8.

Answer 1

Answer:

6.03 m

Explanation:

First of all, let's convert the angular velocity from revolutions per minute to radians per second:

$\omega = 10.9 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=1.14 rad/s$

The frictional force on the block ranges from zero to a maximum value of

$F=\mu mg$

In order for the block to remain stuck on the turntable, the frictional force must be equal to the centripetal force, so we can write:

$m\omega^2 r = \mu mg$

where

m is the mass of the block

$\omega$ is the angular velocity

r is the distance of the block from the centre

$\mu = 0.8$ is the coefficient of static friction

g = 9.8 m/s^2

Solving for r, we find:

$r=\frac{\mu g}{\omega^2}=\frac{(0.8)(9.8)}{(1.14)^2}=6.03 m$