Question

Two particles a and b of mass m are connected by a massless string. A is placed on a rough table. The string passes over a small smooth peg. B is left from a position making an angle 60 with vertical. Minimum coefficient of friction between a and table so that a does not slip during motion of mass b

Answer 1

At a given time, the speed of particle A is 12m/s and that of B is 15m/s. find the total linear momentum of the system of the two particles? ... mass of particle A = 10 g= 0.010 kg. ... mass of particle B = 20 g= 0.020 kg.

Answer 2

The minimum coefficient of friction between A and the table so that A does not slip during the motion of mass B is 1/2.

Given:

• Masses A and B are connected by a massless string.
• A is placed on a rough table.
• B makes an angle of 60° with vertical.

To Find: Minimum coefficient of friction between A and table so that A does not slip during motion of mass B.

Solution:

For A to not slip

Friction = f ≥ T

So, μmg ≥ T

μ ≥ T/mg

μ(min) = T/mg

When B reaches down making 60 degree angle with the verticle.

T = mgcosθ

θ=60

T = mg/2

so μ(min) = T/mg

               = (mg/2)/mg

              = 1/2

Hence minimum coefficient of friction between A and the table so that A does not slip during the motion of mass B is 1/2.

#SPJ3

​