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Hassan Chatterjee
Asked: 2022-11-10T09:12:10+05:30 In:

A circular platform is mounted on a vertical frictionless axle. Its radius is `r = 2 m` and its moment of inertia`I = 200 kg m^2`. It is initially at rest. A `70 kg` man stands on the edge of the platform and begins to walk along the edge at speed `v_0 = 1 m s^-1` relative to the ground. The angular velocity of the platform is.
A. `1.2 rad s^-1`
B. `0.4 rad s^-1`
C. `0.7 rad s^-1`
D. `2 rad s^-1`

A circular platform is mounted on a vertical frictionless axle. Its radius is `r = 2 m` and its moment of inertia`I = 200 kg m^2`. It is initially at rest. A `70 kg` man stands on the edge of the platform and begins to walk along the edge at speed `v_0 = 1 m s^-1` relative to the ground. The angular velocity of the platform is.
A. `1.2 rad s^-1`
B. `0.4 rad s^-1`
C. `0.7 rad s^-1`
D. `2 rad s^-1`
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  1. Correct Answer – C
    ( c) As the system in initially at rest, therefore, initial angular momentum, `L_i = 0`.
    According to the law of conservation of angular momentum, final angular momentum, `L_f = 0`
    :. Angular momentum of man = angular momentum of platform in opposite direction
    i.e., `mv_0 r = I omega`
    or `omega = (m v_0 r)/(I) = (70(1.0)(2))/(200) = 0.7 rad s^-1`.

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Kunti Pradhan
Asked: 2022-11-09T07:30:29+05:30 In:

A circular platform is mounted on a vertical frictionless axle. Its radius is `r = 2 m` and its moment of inertia`I = 200 kg m^2`. It is initially at rest. A `70 kg` man stands on the edge of the platform and begins to walk along the edge at speed `v_0 = 1 m s^-1` relative to the ground. The angular velocity of the platform is.
A. `1.2 rads^(-1)`
B. `0.4rads^(-1)`
C. `2.0rads^(-1)`
D. `0.7rads^(-1)`

A circular platform is mounted on a vertical frictionless axle. Its radius is `r = 2 m` and its moment of inertia`I = 200 kg m^2`. It is initially at rest. A `70 kg` man stands on the edge of the platform and begins to walk along the edge at speed `v_0 = 1 m s^-1` relative to the ground. The angular velocity of the platform is.
A. `1.2 rads^(-1)`
B. `0.4rads^(-1)`
C. `2.0rads^(-1)`
D. `0.7rads^(-1)`
Parabola
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Leave an answer

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1 Answer

  1. Correct Answer – D
    Net external torque is zero. Therefore angular mometnumof system will remain conserved i.e, `L_(i)=L_(f)`
    Initial angular momentum `L_(i)=L_(f)`
    Initial angular momentum `L_(i)=0`
    `:.` final angular mometum should be zero.
    or angular momentum of man `=` angular momentum of platform is opposite direction
    or `mv_(0)r=Iomega`
    `omega=(mv_(0)r)/I=((70)(1.0)(2))/200`
    `omega=0.7rad//s`

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