1. Pulley: Consider a 2.5 kg mass on a horizontal smooth surface attached to a 5

Question

1. Pulley: Consider a 2.5 kg mass on a horizontal smooth surface attached to a 500 g pulley of radius 12 cm which is attached to a hanging mass of 5 kg on a table near the surface of the Earth where the acceleration of gravity is g=9.81 m/s2. Ignore the viscosity of the air.

A. What are the x and y net force equations of motion for each mass, and the net torque equation of motion for the pulley?

B. What is the acceleration of the 5 kg mass?

C. What are the tensions in the rope that act on the pulley?

Explanation / Answer

The only change that comes with pulleys with masses is that they have different on tensions on both sides. Now consider the pulley has tension T1 in vertical section of the rope (5 kg block side) and tension T2 in the horizontal section.

Writing equations for 5 kg block, assuming the system moves downwards

5g - T1 = 5a;

similarly equations for 2.5 kg block are

T2 = 2.5a;

Now there are three unknowns and we have only 2 equations. The 3rd equation comes from the torque equation of the pulley.

(T1 - T2) * r = I * (angular acceleration), where I = mr2 and angular acceleration = linear acceleration / r.

(T1 - T2) * r = (1/2)mr2 * (a / r);

T1 - T2 = (1/2)ma;

as m = (1/2)

T1 - T2 = (1/4)a;

Solving we get a = 20g / 31 = 6.32 m/s2;

T2 = (5/2)a = 15.82N;

T1 = (11/4)a = 17.38N

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