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A solid disc and a ring, both of radius 10cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10pi rad s^(-1). Which of the two will starts to roll earlier ? The co-efficient of kinetic friction is mu_(k)=0.2. |
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Answer» Solution :Frictional force acting on the body, `f=mu_(k)R=mu_(k)mg`………..`(i)` where `R(=mg)` is the NORMAL reaction of the table on the body. If a is the acceleration of the CM of the body `f=ma`…………`(II)` From eqns. (i) and (ii) `ma=mu_(k)mg` or `a=mu_(k)g`............`(iii)` Torque due to frictional force i.e., `tau=fr=(mu_(k)mg)r=mu_(k)mgr` As `tau=Ialpha.alpha=(tau)/(I)=(mu_(k)mgr)/(1)` Linear velocity of CM of the body (initially at rest, `v_(0)=0`) is given by `v=v_(0)+at=at=muk"gt"`.............`(iv)` Angular velocity of the body after time t, i.e. `omega=omega_(0)+alphat=omega_(0)-((mu_(k)mgr)/(I))t`...........(v) (Here `alpha` has been taken NEGATIVE as torque due to frictional force produces retardation) `v=romega`...........(vi) From eqs. (iv), (v) and (vi) we get `mu_(k)"gt"=r[omega_(0)-(mu_(k)mgr)/(I)]t=romega_(0)-((mu_(k)mgr^(2))/(I))t`................(vii) For a ring , as `I=mr^(2)`, from eqn. (vii) `mu_(k)"gt"=romega_(0)-((mu_(k)mgr^(2))/(mr^(2)))t=romega_(0)-mu_(k)"gt"` or `2mu_(k)"gt"=romega_(0)` or `t=(romega_(0))/(2mu_(k)g)` or `t=(0.1xx10pi)/(2xx0.2xx9.8)=0.80s` (as `r=0.1omega_(0)=10pirad//s`, `mu_(k)=0.2`, `g=9.8m//s^(2)`) or `3mu_(k)"gt"=romega_(0)` or `t=(romega_(0))/(3mu_(k)g)` or `t=(0.1xx10pi)/(3xx0.2xx9.8)=0.53s` Since t is less in case of disc the disc begin to roll earlier than the ring for the same VALUES of r and `omega_(0)` |
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