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TWO BOYS ARE RIDING ON A MERRY GO ROUND WHICH ARE TURNING AT A CONSTANT RATE.BOY 'A' IS AT OUTER RIM.AND BOY 'B' IS CLOSER TO CENTER OF ROTATION.WHO HAS TO HOLD ON TIGHTER?PLZZZ EXPLAIN WITH REASON.(ASSUME THEIR MASSES TO BE SAME).............??????????????
boy B needs hold more tightly because 1. boy B has to face more centrifugal force as mv2/r=F 2. Boy B has less moment of inertia(mr2)
Boy A has to hold tightly. Reason: We know, Torque = RxF Hence, we can say, that the person who is at a greater distance from the axis of rotation (center of merry-go-round) experiences a greater torque. Since, he faces more torque, hence boy A needs to apply greater force to balance that torque. Therefore, Boy A needs to hold tighter. Please Approve!!
Boy A has to hold tightly.
Reason:
We know,
Torque = RxF
Hence, we can say, that the person who is at a greater distance from the axis of rotation (center of merry-go-round) experiences a greater torque. Since, he faces more torque, hence boy A needs to apply greater force to balance that torque. Therefore, Boy A needs to hold tighter.
Please Approve!!
torque = F*r A F=mv2/r T=mv2/r *r=mv2 B F=mv2/R T=mv2/R *R=mv2 (mv2/r is centrifugal force) So, A and B experience the same torque B experiences greater force as mv2/r=F So r of B is smaller than A so, force will be greater.........
pls cancel all the ans above , the correct ans is- Boy A because it has to face the centrifugal force mv2/R = m(wr)2/r=mw2r. w is equel here so the one with greater radius will face more force , hencce boy A has to hold more tightly. THIS IS CORRECT!!!!! Sorry for wrong ans above.
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