Question

a heavy solid sphere is thrown on a horizontal rough surface with initial velocity 7 m/s without rolling.when it starts pure rolling motion,then its speed will be??

Answer 1

Answer on Question#39726, Physics, Mechanics

Question:

A heavy solid sphere is thrown on a horizontal rough surface with initial velocity $7\,\mathrm{m/s}$ without rolling. When it starts pure rolling motion, then its speed will be??

Solution:

For translational motion we have

$$ma = -\mu mg$$

where $\mu$ is the coefficient of friction

So

$$a = -\mu g$$

After time $t$ velocity is

$$v = v_0 + at = v_0 - \mu gt$$

For rotational motion about centre a net torque acting upon an object will produce an angular acceleration of the object according to

$$\tau = l\alpha$$ $$\mu mgr = l\alpha$$

where $r$ is radius of sphere, $l$ is moment of inertia, $\alpha$ is angular acceleration

The moment of inertia of solid sphere of radius $r$ and mass $m$ is

$$I = \frac{2mr^2}{5}$$

So

$$\mu mgr = \frac{2mr^2}{5}\alpha$$ $$\alpha = \frac{5\mu g}{2r}$$

The angular velocity is

$$\omega = 0 + \alpha t$$ $$\omega = \frac{5\mu g}{2r}t$$

For pure rolling motion

$$v = r\omega$$

So

$$v_0 - \mu gt = r \frac{5\mu g}{2r}t$$ $$v_0 = \mu gt + \frac{5\mu gt}{2} = \frac{7\mu gt}{2}$$

From this equation

$$t = \frac{2v_0}{7\mu g}$$

Now we put $t$ in equation for $v$

$$v = v_0 - \mu gt = v_0 - \frac{\mu g 2v_0}{7\mu g} = v_0 - \frac{2v_0}{7} = \frac{5v_0}{7}$$ $$v = \frac{5 \cdot 7}{7} = 5\,\mathrm{m/s}$$

Answer. $v = 5\,\mathrm{m/s}$