InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A solid disc and a ring, both of radius `10 cm` are placed on a horizontal table simultaneously, with initial angular speed equal to `10 pi rad//s`. Which of the two will start to roll earlier ? The coefficient of kinetic friction is `mu_(k) = 0.2`. |
|
Answer» Here, initial velocity of centre of mass is zero i.e. `u = 0`. Frictional force causes the c.m to accelerate. `mu_(k) xx mg = ma :. A = mu_(k)g` …(i) As `v = u + at :. v = 0 + mu_(k) g t` ..(ii) Torque due to friction causes retardation in the initial angular speed `omega_(0)`. i.e. `mu_(k) mg xx R = - I alpha` `alpha = (-mu_(k) mg R)/(I)` ..(iii) As `omega = omega_(0) + alpha t :. omega = omega_(0) - (mu_(k) mg R t)/(I)` ...(iv) Rolling begins, when `v = R omega` From (ii) and (iv) `mu_(k) g t = R omega_(0) - (mu_(k)mg R^(2)t)/(I)` ..(v) `:. mu_(k) g t = R omega_(0) - mu_(k) g t :. omega_(0) = (2mu_(k) g t)/(R ) or t = (omega_(0)R)/(2mu_(k) g)` ..(vi) For a disc, `I = (1)/(2)mR^(2)` `:.` from (v), `mu_(k) g t = R omega_(0) - 2 mu_(k) g t` `2mu_(k) g t = R omega_(0)` `t = (omega_(0)R)/(2mu_(k) g)` ..(vii) Comparing (vi) and (vii), we find that the disc would begin to roll earlie than the ring. We can calculate the values of `t` from (vi) and (vii) using known values of `mu_(k), g, R` and `omega_(0)`. |
|