Consider two masses of 3.2 kg and 7.6 kg connected by a string passing over a pu
ID: 2299568 • Letter: C
Question
Consider two masses of 3.2 kg and 7.6 kg connected by a string passing over a pulley having a moment of inertia 9.1 g . m2 about its axis of rotation, as in the figure below. The string does not slip on the pulley, and the system is released from rest. The radius of the pulley is 0.24 m. Find the linear speed of the masses after the 7.6 kg mass descends through a distance 25 cm. Assume mechanical energy is conserved during the motion. The acceleration of gravity is 9.8 m/s2 . Answer in units of m/sExplanation / Answer
m1 = 7.6........m2 = 3.2
for m1
m1*g-T1 = m1*a
T1 = m1*g - m1*a......(1)
for m2
T2 - m2g = m2*a
T2 = m2*a + m2*g........(2)
for pulley
(T1 - T2)*R = I*alfa
(m1*g - m1*a - m2*a - m2*g)*R = I*a/R
m1*g - m2*g - a*(m1+m2) = I*a/R^2
(m1-m2)*g - a*(m1+m2) = I*a/R^2
(m1-m2)*g = a*((m1+m2) + I/R^2 )
a = 3.94 m/s^2
V^2 -u^2 = 2*a*s
v^2 = 2*3.94*0.25
V = 1.4 m/s