# A circular platform is mounted on a vertical frictionless axle.Its radius is 2m and its moment of inertia is 200 .. its initially at rest.a 70 kg man stands on the edge of the platform and begins to walk along the edge at speed v =1 m/s relative to the ground .Angular velocity of the platform is??

Aman Bansal
592 Points
12 years ago

Dear Sai,

Applying law of conservation of angular momentum in vertical direction (axis of rotation), we have :

Li=Lf

Both platform and the girl are initially stationary with respect to ground. As such, initial angular momentum of the system is zero. From conservation law, it follows that the final angular momentum of the system is also zero.
But final angular momentum has two constituents : (i) final angular momentum of the platform and (ii) final angular momentum of the girl. Sum of the two angular momentums should, therefore, be zero. Let “P” and “G” subscripts denote platform and girl respectively, then :

Lf=LPf+LGf=0

Now, MI of the circular platform about the axis of rotation is :

IP=MR22=100x322=450kg−m2

The mass of the particles constituting the girl are at equal perpendicular distances from the axis of rotation. Its MI about the axis of rotation is :

IG=mR2=60x32=540kg−m2

From the question, it is given that final speed of the girl around the rim of the platform is 1 m/s in clockwise direction. It means that girl has linear tangential velocity of 1 m/s with respect to ground. Her angular velocity, therefore, is :

Negative sign indicates that the girl is rotating clockwise. Now, putting these values, we have :

LPf+LGf=IPωPf+IGωGf=0

The positive value of final angular velocity of platform means that the platform rotates in anti-clockwise direction.
Note : Alternatively, we can find angular momentum of the girl, using the defining equation for angular momentum of a point (considering girl to be a particle like mass) as :

LGf=−mvr=−60x1x3=−180kg−ms2

Best Of luck

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Thanks

Aman Bansal