Rotational Motion

JEE Main level, numerical answer, medium

55 of 280

Question 55 of 280

A solid disc of radius 20cm20cm and mass 10kg10kg is rotating with an angular velocity of 600rpm600rpm, 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 10s10s is _________π×10−1Nm\pi \times {10}^{-1}Nm

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Answer: 4

τ=ΔLΔt=I(ωf−ωi)Δt\tau =\dfrac{\Delta L}{\Delta t}=\dfrac{I\left({\omega }_{f}-{\omega }_{i}\right)}{\Delta t}
τ=mR22×[0−ω]Δt\tau =\dfrac{\dfrac{m{R}^{2}}{2}\times [0-\omega ]}{\Delta t}
=10×(20×10−2)22×600×π30×10=\dfrac{10\times {\left(20\times {10}^{-2}\right)}^{2}}{2}\times \dfrac{600\times \pi }{30\times 10}
=0.4π=4π×10−1=0.4\pi =4\pi \times {10}^{-1}

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