A uniform solid sphere of mass and radius is given an initial linear velocity without rotation on a rough horizontal surface. The coefficient of kinetic friction between sphere and ground is . 🔍 Find: Time when pure rolling starts. Distance travelled before pure rolling.

1. Forces and torques acting on the sphere

• A kinetic-friction force $f_k = \mu_k \, m g$ acts opposite to motion.
• This force simultaneously produces:
  • Linear deceleration $a = -\mu_k g$.
  • Torque $\tau = f_k R$ about the centre, giving angular acceleration.

2. Equations of motion (with symbols)

• Translational motion $\frac{dv}{dt} = -\mu g \;\;\Rightarrow\;\; v(t) = v_0 - \mu g \, t$
• Rotational motion
  • Moment of inertia of a solid sphere: $I = \frac{2}{5} m R^2$
  • Angular acceleration: $\alpha = \frac{\tau}{I} = \frac{f_k R}{I} = \frac{\mu m g R}{\tfrac{2}{5} m R^2} = \frac{5}{2}\,\frac{\mu g}{R}$
  • Angular speed: $\omega(t) = \alpha t = \frac{5}{2}\,\frac{\mu g}{R}\, t$

3. Pure rolling condition

For rolling without slipping: $v = \omega R$ Plug $v(t)$ and $\omega(t)$, solve for time $t_r$ when equality first occurs.

4. Distance covered till that instant

Use constant-acceleration kinematics: $s = v_0 t_r + \tfrac{1}{2} a t_r^2$ Insert the value of $t_r$ and $a = -\mu g$ to find $s$.

Why each step?

• Step 1: Identify forces → needed to write Newton’s laws.
• Step 2: Separate translation and rotation → each obeys its own equation but both share the same friction.
• Step 3: Apply the rolling criterion → gives the exact instant sliding stops.
• Step 4: Classic kinematics → distance under uniform acceleration/deceleration.

What’s happening?

Imagine you push a smooth cricket ball on a dusty floor. At first, the ball slides without spinning. The rough floor rubs it (friction), so:

1. It slows down (because friction opposes the slide).
2. That same friction twists the ball, making it spin.

After some time the ball’s forward speed and spin match perfectly so that the bottom point of the ball is momentarily at rest with the floor – that’s called pure rolling (no sliding at all, only rolling like a wheel). We need to know:

• How long it takes to reach that perfect rolling.
• How far the ball travels before it reaches that state.

Key ideas in baby language

• Friction is like a hand: it pulls back on the ball (slowing it) and at the same time twists it (spins it).
• The ball has to obey the condition for rolling: speed = spin × radius.
• We’ll write simple maths to find the time when this magic equality happens and how much ground the ball covered till then.

Step-by-step solution

1. Linear (translational) motion
   $f_k = \mu m g \;\;(\text{opposite to }v_0)$
   $a = -\mu g$
   $v(t) = v_0 - \mu g t$

2. Rotational motion
   $I = \frac{2}{5} m R^2$
   $\tau = f_k R = \mu m g R$
   $\alpha = \frac{\tau}{I} = \frac{\mu m g R}{\tfrac{2}{5} m R^2} = \frac{5}{2}\,\frac{\mu g}{R}$
   $\omega(t) = \alpha t = \frac{5}{2}\,\frac{\mu g}{R}\, t$

3. Condition for pure rolling
   $v(t_r) = \omega(t_r) R$
   $v_0 - \mu g t_r = \left(\frac{5}{2}\,\frac{\mu g}{R} t_r\right) R = \frac{5}{2}\,\mu g t_r$
   $v_0 = \mu g t_r \left(1 + \frac{5}{2}\right) = \mu g t_r \cdot \frac{7}{2}$
   $\boxed{t_r = \frac{2 v_0}{7 \mu g}}$

4. Distance covered until rolling begins
   $s = v_0 t_r + \tfrac{1}{2} a t_r^2$
   Substitute $a = -\mu g$ and $t_r$:

   $s = v_0 \left(\frac{2 v_0}{7 \mu g}\right) - \frac{1}{2} \mu g \left(\frac{2 v_0}{7 \mu g}\right)^2$
   $s = \frac{2 v_0^2}{7 \mu g} - \frac{1}{2} \mu g \cdot \frac{4 v_0^2}{49 \mu^2 g^2}$
   $s = \frac{2 v_0^2}{7 \mu g} - \frac{2 v_0^2}{49 \mu g}$
   $s = \frac{2 v_0^2}{\mu g}\left(\frac{1}{7} - \frac{1}{49}\right) = \frac{2 v_0^2}{\mu g}\left(\frac{6}{49}\right)$
   $\boxed{\displaystyle s = \frac{12}{49} \; \frac{v_0^2}{\mu g}}$

Final Answers

• Time to start pure rolling: $t_r = \dfrac{2 v_0}{7 \mu g}$
• Distance travelled before pure rolling: $s = \dfrac{12 v_0^2}{49 \mu g}$