Question

Initially, the monkey is standing on the floor and the bananas are hanging above his head near the pulley. The monkey’s mass is 2 kg and the mass of the bananas is 4 kg. The pull is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a constant torque of magnitude 0.50 N. If the bananas and the monkey are released from rest with the bananas 1.0 m off the floor, how long does it take the bananas to reach the ground?

Answer #1

Newton's second law : F = ma. and Tr = I

Linear for bananas

T1 - 4g = -4a

T1 + 4a = 40 ...i

Linear for monkey

T2-2g = 2a

T2-2a = 20 ..... ii

Angular for pulley

T1 - T2 -Tf = I

T1x0.06 - T2x0.06 - 0.5 = (2 x 0.06^2 / 2) x

0.06T1 - 0.06T2 -0.0036 = 0.5

but = a/r = a/0.06

0.06T1 - 0.06T2 - 0.06a = 0.5 .... iii

solving 3 eqns

T1 = 33.33N

T2 = 23.33N

a = 1.67 m/s^2

using second kinematical equations

s = 1/2 a t^2

t^2 = 2s/a = 2x1/1.67 = 1.2

**t = 1.09 s**

**upvote if helps**

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