Question

A mass of 150g is suspended vertically from a string that runs over a pulley and attached to a 100g mass on a rough horizontal surface. The pulley has a mass of 30g and a radius of 3.5cm. The string moves without slipping over the pulley while the masses move. The coefficient of kinetic friction between the rough surface and the 100-g mass is 0.05. Calculate the acceleration of the masses.

Answer 1

M = 150 g = 0.15 kg
m = 100 g = 0.1 kg
mp = 30 g = 0.03 kg
r = 3.5 cm = 0.035 m
uk = 0.05
I = 1/2 * mp * r^2
Let the tension in the string attache to 150g mass = T1
Let the tension in the string attache to 100g mass = T2

M*g - T1 = M*a
0.15*9.8 - 0.15 * a = T1

T2 - uk*m*g = m*a
T2 = 0.1*a + 0.05*0.1*9.8

(T1 - T2)*r = 1/2 * I * (a/r)
(0.15*9.8 - 0.15 * a - (0.1*a + 0.05*0.1*9.8 ) = 1/2 * 1/2 * mp * r^2 * a/r^2
(0.15*9.8 - 0.15 * a - (0.1*a + 0.05*0.1*9.8 ) = 1/4 * 0.03 * a
a = 5.52 m/s^2
Acceleration, a = 5.52 m/s^2