Aman Bansal
Last Activity: 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 :
ωGf=−vR=−13rad/s
Negative sign indicates that the girl is rotating clockwise. Now, putting these values, we have :
LPf+LGf=IPωPf+IGωGf=0
ωPf=−−1x5403x450=0.4rad/s
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
Askiitian Expert
