A disc is standing on a flat rough surface. Its centre is suddenly given a velocity of 3m/s in the forward direction. In how much time will pure rolling start? Mass = 1kg, Radius = 10 cm. Coefficient of friction = 0.4. |
0.05 s 0.25 s 1 s 0.5 s |
0.25 s |
Friction force will decrease its linear velocity & increase its angular velocity. f = μN = μmg = 4N = ma ⇒ $a = 4 m/s^2$ After time t its velocity will be $V = V_0 - at = 3 - 4t$ Angular velocity a time t is given by $\omega = \omega_0 +\alpha t = \alpha t$ where $\alpha = \frac{\tau}{I} = \frac{fr}{mr^2/2} = \frac{2f}{mr} = \frac{2\times 4}{1\times 0.1} = 80 rad/s^2$ At the time of pure rolling $V = \omega r$ $ \Rightarrow 3 - 4t = \alpha r t = 8t$ $\Rightarrow t = \frac{3}{12} = 0.25$ |