Question

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 3.30 kg) moves is frictionless. The system is released from rest, and block B (mass 6.00 kg ) moves downward 1.80 m m in 2.00 s

A) What is the tension force that the rope exerts on block B?

B) What is the tension force that the rope exerts on block A?

C) What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

Answer 1

Given the mass of block A is mA = 3.30kg, the mass of B is m = 6kg, the radius of the pulley is R = 0.080m and the displacement of the block B is h = 1.80m in t = 2s. The forces acting on the system is shown below.

Since the system is at rest initially, the initial speed of the block B is u = 0m/s. By the equation of motion,

So the acceleration of the system is 0.9m/s^2.

A) From the figure, the second law equation for the block B is,

So the tension force that the rope exerts on block B is 53.4N.

B) The second law equation for the block A is,

So the tension force that the rope exerts on block A is 2.97N.

C) If I is the moment of inertia of the pulley and is its angular acceleration, then the torque acting on the pulley is given by,

But,

Also,

Therefore,

So the moment of inertia of the pulley is 0.359kgm^2.