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A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simulataneously, with initial angular speed equal to 10 pi rad s^(-1). Which of the two will starts to roll earlier ? The co - efficient of kinetic fricition 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 = I alpha.alpha=(tau)/(I)=(mu_(k)mgr)/(1)` Linear velocity of CM of the body (initially at rest, `upsilon_(0)=0`) is give by `upsilon = upsilon_(0)+at = at = mu kgt ""`....(iv) Angular velocity of the body after time t, i.e., `omega = omega_(0)+alpha t = omega_(0)-((mu_(k)mgr)/(I))t ""`......(v) (Here, `alpha` has been taken negative as torque due to frictional force produces retardation.) `upsilon = r omega ""`.....(vi) From eqns.(iv), (v) and (vi), we get `mu_(k)g t = r [omega_(0)-((mu_(k)mgr)/(I))t]=r omega_(0)-((mu_(k)mg r^(2))/(I))t ""`....(VII) For a RING, as `I = mr^(2)`, from eqn. (vii) `mu_(k) g t=r omega_(0)-((mu_(k)mgr^(2))/(mr^(2)))t=r omega_(0)-mu_(k)g t` or `2mu_(k)g t=r omega_(0)` or `t = (r omega_(0))/(2 mu_(k)g)` or `t = (0.1xx10pi)/(2xx0.2xx9.8)=0.80s` `("as " r = 0.1 omega_(0)=10 pi rad//s, mu_(k)=0.2, g=9.8 m//s^(2))` or `3 mu_(k)g t = r omega_(0)` or `t=(r omega_(0))/(3mu_(k)g)` or `t=(0.1xx10 pi)/(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 value of r and `omega_(0)`. |
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