A pulley of radius 2 m is rotated about its axis by a force F = 20t - 5t^2 Newton (where t is measured in seco

Question

A pulley of radius 2 m is rotated about its axis by a force F = 20t - 5t^2 Newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m^2 , the number of rotations made by the pulley before its direction of motion is rever

Accepted Answer

1. Calculate Torque and Angular Acceleration:

τ = F \times r = (20 t - 5 t^{2}) \times 2 = 40 t - 10 t^{2}

α = \frac{τ}{I} = \frac{40 t - 10 t ^{2}}{10} = 4 t - t^{2}

2. Find time

t

when velocity reverses (

ω = 0

):

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

Set

ω = 0

:

2 t^{2} - \frac{t ^{3}}{3} = 0 ⟹ t = 6 s

.3. Find angular displacement

θ

:

θ = \int_{0}^{6} ω d t = \int_{0}^{6} (2 t^{2} - \frac{t ^{3}}{3}) d t = [\frac{2 t ^{3}}{3} - \frac{t ^{4}}{12}]_{0}^{6}

θ = \frac{2 ( 216 )}{3} - \frac{1296}{12} = 144 - 108 = 36 rad

4. Number of rotations

n

:

n = \frac{θ}{2 π} = \frac{36}{2 π} \approx \frac{36}{6.28} \approx 5.73

This is more than 3 but less than 6.

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