Question

Two blocks are connected by a light string that passes over a frictionless pulley having a moment of inertia of 0.0040 kg*m2 and a radius of 5.0 cm. The coefficient of kinetic friction between the table top and the upper block is 0.300. The blocks are released from rest. Using energy methods, find the speed of the upper block just as it has moved 0.600 m.

Answer 1

Let’s find the linear acceleration first, and let us find what the linear acceleration of the pulley is.

Since there is no slipping, the linear acceleration of the pulley equals the linear acceleration of the whole system.

Let's find the net torque on the pulley:

The net force on the pulley is the tension in the vertical line T1 minus the tension in the horizontal line, T2.

The tension in the vertical line is

T1 = mg - ma = 2g - 2a where a the acceleration of the system is.

and T2 is

T2 = mg? + ma = 3g*0.3 + 3a where mg? the friction force is.

the net force on the pulley is then

Fnet = T1 - T2 = 2g - 2a - 3a - 3*0.3g

Fnet = 1.1g - 5a

and the torque on the pulley is then

? = RF = I*? = I*a/R with ? = a/R (? = angular acceleration, I = moment of inertia, R = 0.05 m)

(1.1g - 5a)*0.05 = 0.004*a/0.05

(1.1g - 5a)*0.05^2 = 0.004a

1.1*9.81*0.05^2 - 0.05^2*5a = 0.004a

a = 1.635 m/s^2

Since we know a now, we can find v with

v = srrt(2as)

v = sqrt(2*1.635*0.6)

v = 1.4007 m/s <--- ans.