Question

A man of mass 50kg is moving in a circular path of radius R=1m with speed v varying with time as V=kt, where t is in seconds and k=1/3m/s2.uf coefficient of friction between ground and man is u=.1 then the time at which sliding starts,is

Answer 1

Answer:

The time at which the sliding will start is 3 seconds.

Explanation:

Given

Mass of the man = 50 kg

Radius of the circular path R = 1 m

Speed v = kt

Where k = 1/3 m/s²

Therefore, v = t/3 m/s

v is the tangential velocity

The tangential acceleration acting on the man

a = dv/dt

  = $\frac{d}{dt}(t/3)$

  = 1/3 m/s²

Force acting in the tangential direction

F₁ = ma = 50 × 1/3 = 16.67 N

The frictional force

F = μN

  = 0.1 × 50 × 10 N

 = 50 N

Which is greater than the Force acting in the tangential direction

Thus, the sliding will occur only when the centrifugal force will be equal to the frictional force

Centrifugal force = mv²/R

The sliding will occur when

mv²/R = μN

⇒ mv²/R = μmg

⇒ v²/R = μg

⇒ (t/3)²/1 = 0.1 × 10

⇒ t² = 9

⇒ t = 3

Therefore, the sliding will start after 3 seconds.