JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Rotational Mechanics

A pulley of radius 1.5 m is rotated about its axis by a force

F = (12 t - 3 t^{2}) N

applied tangentially (while

t

is measured in seconds). If moment of inertia of the pulley about its axis of rotation is

4.5 kg m^{2}

, the number of rotations made by the pulley before its direction of motion is reversed, will be

\frac{K}{π}

. The value of

K

____

Answer: 18

Solution

The torque is

τ = F r = (12 t - 3 t^{2}) \times 1.5

. Using

τ = I α

α = \frac{τ}{I} = \frac{( 12 t - 3 t ^{2} ) \times 1.5}{4.5} = 4 t - t^{2}

Integrating to find angular velocity:

ω = \int_{0}^{t} (4 t - t^{2}) d t = 2 t^{2} - \frac{t ^{3}}{3}

The direction reverses when

ω = 0

2 t^{2} - \frac{t ^{3}}{3} = 0 \Rightarrow t^{2} (2 - \frac{t}{3}) = 0

This gives

t = 0

t = 6

s.

Integrating

ω

to find angular displacement:

θ = \int_{0}^{6} (2 t^{2} - \frac{t ^{3}}{3}) d t = [\frac{2 t ^{3}}{3} - \frac{t ^{4}}{12}]_{0}^{6} = \frac{2 \times 216}{3} - \frac{1296}{12} = 144 - 108 = 36

rad

Number of rotations:

n = \frac{θ}{2 π} = \frac{36}{2 π} = \frac{18}{π}

Therefore,

K = 18

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer