InterviewSolution
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InterviewSolution
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A ring of radius `R` is rotating with an angular speed `omega_0` about a horizontal axis. It is placed on a rough horizontal table. The coefficient of kinetic friction is `mu_k`. The time after it starts rolling is.A. `(omega_0 mu_k R)/(2 g)`B. `(omega_0 g)/(2 mu_k R)`C. `(2 omega_0 R)/(mu_k g)`D. `(omega_0 R)/(2 mu_k g)`] |
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Answer» Correct Answer - D (d) Acceleration produced in the centre of mass due friction. `a = (f)/(M) = (mu_k Mg)/(M) = mu_k g` where `M` is the mass of the ring Angular retardation produced by the torque due friction `prop = (tau)/(I) = (f R)/(I) = (mu_k M g R)/(I)` As `v = u + at` :. `v = 0 + mu k gt (because u = 0)` (Using (i)) As `omega = omega_0 + prop t` :. `omega = omega_0 - (mu_k M gR_1)/(I)` (Using (ii)) m For rolling without slipping `v = R omega` :. `(v)/(R) = omega_0 - (mu_k Mg T_1)/(I) t` `(mu_k gt)/(R) = omega_0 - (mu_k M gR)/(I) t rArr (mu_k gt)/(R) [1 + (MR^2)/(1)] =omega` `(mu_k gt)/(R) = (omega_0)/(1+ (MR^2)/(I)) rArr t = (R omega_0)/(mu_k g((1 + MR^2)/(I)))` For ring, `I = MR^2` :. `t = (R omega_0)/(mu_k g(1 + (MR^2)/(MR^2)))=(R omega_0)/(2 mu_k g)`. |
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