JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Rotational Mechanics

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A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in 10s is _____

π \times 1 0^{- 1}

Nm.

Answer: 4

Solution

The moment of inertia of the disc about an axis through its centre and normal to its plane is

I = \frac{1}{2} M R^{2} = \frac{1}{2} \times 10 \times (0.2)^{2} = 0.2 kg m^{2}

The initial angular speed is

ω = 600

rpm

= \frac{600 \times 2 π}{60} = 20 π

rad/s, and it falls to zero in

10

s, so the magnitude of the angular retardation is

α = \frac{20 π}{10} = 2 π

rad/s

^{2}

. The torque needed is

τ = I α = 0.2 \times 2 π = 0.4 π N m

The retarding torque is

4 π \times 1 0^{- 1}

N m.

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Rotational Mechanics

Solution

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