Question

a man of mass 50 kg is moving in a circular path of radius R= 1 metre with speed v varyng with time as v= kt where t is a time in second and k equal to 1 by 2 metre per second square if the coefficient of friction between the ground and the man is .1 then the time taken at which it sliding start

Answer 1

Answer:

The time at which the sliding will start is 2 seconds.

Explanation:

Given

Mass of the man = 50 kg

Radius of the circular path R = 1 m

Speed v = kt

Where k = 1/2 = 0.5 m/s²

Therefore, v = 0.5t m/s

v is the tangential velocity

The tangential acceleration acting on the man

a = dv/dt

  = $\frac{d}{dt}(0.5t)$

  = 0.5 m/s²

Force acting in the tangential direction

F₁ = ma = 50 × 0.5 = 25 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 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

⇒ (0.5t)²/1 = 0.1 × 10

⇒ t² = 1/0.25

⇒ t² = 100/25

⇒ t = 10/5 = 2 seconds

Therefore, the sliding will start after 2 seconds.

Answer 2

Given Mass of the man = 50 kg

Radius of the circular path R = 1 m

Speed v = kt where k = 1/2 = 0.5 m/s²

Therefore, v = 0.5 t m/s v is the tangential velocity.

The tangential acceleration acting on the man a = dv/dt  = 0.5 m/s²

Force acting in the tangential direction F₁ = ma = 50 × 0.5 = 25 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 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

⇒ (0.5t)²/1 = 0.1 × 10

⇒ t² = 1/0.25

⇒ t² = 100/25

⇒ t = 10/5 = 2 seconds

Therefore, the sliding will takes place at 2 secs.