First slide

Dynamics of rotational motion about fixed axis

Question

A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2 $π$ revolutions is [NEET 2019]

Moderate

A

$2 \times 10^{- 3} N - m$
B

$12 \times 10^{- 4} N - m$
C

$2 \times 10^{6} N - m$
D

$2 \times 10^{- 6} N - m$

Solution

Given, mass of cylinder, $m = 2 kg$
Radius of cylinder, $r = 4 cm = 4 \times 10^{- 2} m$
Rotational velocity, $3 rpm = 3 \times \frac{2 π}{60} = \frac{π}{10} {rads}^{- 1} and$
$θ = 2 π \times revolution = 2 π \times 2 π = 4 π^{2} rad$
The work done in rotating an object by an angle $θ$ from rest is given by $W = τ θ$
As the cylinder is brought to rest, so the work done will be negative.
According to work-energy theorem,
Work done = Change in rotational kinetic energy
$- τ θ = \frac{1}{2} I ω_{f}^{2} - \frac{1}{2} I ω_{i}^{2} = \frac{1}{2} I (ω_{f}^{2} - ω_{i}^{2})$
$\Rightarrow τ = \frac{I (ω_{i}^{2})}{2 θ}$ $(∵ ω_{f} - 0)$
$= \frac{1}{2} (\frac{1}{2} m r^{2}) \frac{ω_{i}^{2}}{θ} [∵ I = \frac{1}{2} m r^{2} (for cylinder)]$
$= \frac{1}{4} m r^{2} \frac{ω^{2}}{θ} (∵ ω_{i} = ω)$
$= \frac{1}{4} \times 2 \times {(4 \times 10^{- 2})}^{2} \times {(\frac{π}{10})}^{2} \times \frac{1}{4 π^{2}}$
$= \frac{1}{4} \times 2 \times 16 \times 10^{- 4} \times \frac{π^{2}}{100} \times \frac{1}{4 π^{2}}$
$= \frac{2}{100} \times 10^{- 4} = 2 \times 10^{- 6} N - m$

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