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Standard XII

Physics

Kinetic Energy of a Rigid Body

A rotating pl...

Question

A rotating platform of moment of inertia $100$ kg- $m^{2}$ completes one rotation in $10$ seconds A person of mass is $50 k g$ is standing at the center of the platform. If a person moves at a distance of $2 m$ from the center along the radius then find the angular velocity of the platform.

\frac{π}{30} r a d / s e c

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\frac{π}{60} r a d / s e c

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\frac{π}{15} r a d / s e c

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\frac{π}{20} r a d / s e c

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Solution

The correct option is B $\frac{π}{15} r a d / s e c$

Given,

The platform completes one rotation in 10 sec.

$ω = \frac{2 π}{10} = \frac{π}{5}$

Initially $I = 100 k g - m^{2}$

Finally $I = 100 + 50 \times 2^{2} = 300 k g - m^{2}$

In this process of changing I, no external torque is involved,

Hence,

$I_{1} ω_{1} = I_{2} ω_{2}$

100\times \dfrac{\pi}{5}=300\omega_2$

$ω_{2} = \frac{π}{5 \times 3} = \frac{π}{15}$

Option $C$ is the correct answer

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A rotating platform of moment of inertia 100 kg-m2 completes one rotation in 10 seconds A person of mass is 50kg is standing at the center of the platform. If a person moves at a distance of 2m from the center along the radius then find the angular velocity of the platform.

A rotating platform of moment of inertia $100$ kg- $m^{2}$ completes one rotation in $10$ seconds A person of mass is $50 k g$ is standing at the center of the platform. If a person moves at a distance of $2 m$ from the center along the radius then find the angular velocity of the platform.