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!!
 
						

							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.