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 $2\pi$ revolutions is

• A. $2\times {10}^{-6}Nm$
• B. $2\times {10}^{-3}Nm$
• C. $12\times {10}^{-4}Nm$
• D. $2\times {10}^{6}Nm$

Answer 1

The correct option is A $2\times {10}^{-6}Nm$

θ = 2π×2π = 4π2θ = 2π×2π = 4π^2θ = 2π×2π = 4π2

ω0=3 rpm =3 2π/60 rad/sω_0 =3 rpm =3 2π/60 rad/sω0​=3 rpm =3 2π/60 rad/s

ω2= ω02 − 2αθω^2 = ω_0^2 - 2αθω2= ω02​ − 2αθ ⇒ 0 = (3×2π60)2− 2α4π20 = (3×2π60)^2 - 2α4π^20 = (3×2π/60)^22− 2α4π2

$\alpha =-\frac{1}{800}$ $\frac{rad}{se{c}^2}$

$torque=I\alpha =\left(\frac{16}{{10}^4}\right)(-\frac{1}{800})=-2\times {10}^{-6}Nm$