Question

A metal block of mass 235 g rests at a point 2.8 m from the center of a horizontal rotating wooden platform. The coefficient of static friction between the block and the platform is 0.437. The platform initially rotates very slowly but the rotation rate is gradually increasing. The acceleration of gravity is 9.8 m/s 2 . At what minimum angular velocity of the platform would the block slide away? Answer in units of rad/s.

Answer 1

Answer:

ω=1.23 rad/sec

Explanation:

The block will start sliding when the centrifugal force exceeds the force of friction

We have Force of friction and centrifugal force are as under

$F_{f}=\mu mg...........(i)\\\\F_{c.f}=m\omega ^{2}r................(ii)$

Thus equating the 2 forces to get the minimum angular velocity we have

$\fr\mu mg= m\omega^{2}r\\\\\therefore \omega =(\frac{\mu g}{r})^{1/2}$

Applying values we get

ω=1.23 rad/sec