A uniform disc of mass m and radius R is projected horizontally with velocity v_0 on a rough horizon

Q: Why does friction act backwards?

Because the disc is initially purely translating, the bottommost point is slipping forward relative to the ground. Kinetic friction always opposes relative slipping, so it acts backwards.

Q: Why does friction stop doing work after $t_0$?

At $t = t_0$, pure rolling begins. In pure rolling, the point of contact is instantaneously at rest relative to the ground. Since there is no relative displacement at the point of application of the force, the work done by static friction is zero.

Q: Can we calculate work done by integrating force over distance?

Yes, but it's much more complicated! You would have to integrate the friction force over the actual displacement of the *bottommost point* of the disc, not the center of mass. The Work-Energy Theorem is a much more elegant and less error-prone approach.

Q: Why is the angular acceleration positive?

Friction acts backwards at the bottom of the disc. If you take the torque about the center of mass, this force creates a torque that tends to rotate the disc in the forward (rolling) direction. We define this direction as positive for $\omega$ and $\alpha$.

Q: Does the total mechanical energy remain conserved?

No, not during the slipping phase ($t t_0$), mechanical energy is conserved.

Question

A uniform disc of mass $m$ and radius $R$ is projected horizontally with velocity $v_0$ on a rough horizontal floor, so that it starts off with a purely sliding motion at $t = 0$. After $t_0$ seconds, it acquires a purely rolling motion as shown in figure.
(a) Calculate the velocity of the centre of mass of the disc at $t_0$.
(b) Assuming the coefficient of friction to be $\mu$, calculate $t_0$. Also calculate the work done by the frictional force as a function of time and the total work done by it over a time $t$ much longer than $t_0$.

Sigma Percentile · Accepted Answer

Imagine a uniform disc sliding onto a rough horizontal floor. Initially, it's just translating with a velocity $v_0$—no rotation at all. This is a classic scenario in rotational dynamics where a purely translating body eventually transitions into pure rolling due to the action of kinetic friction.

### Analyzing the Setup

At $t = 0$, the disc has a linear velocity $v_0$ and an angular velocity $\omega = 0$. Because the bottommost point of the disc is scraping forward against the floor with velocity $v_0$, kinetic friction immediately kicks in. This friction force, $f_k = \mu mg$, acts backwards to oppose the slipping.

This single force performs two crucial tasks simultaneously:
1. **Linear Retardation:** It slows down the forward translation. The retardation is $a = \frac{f_k}{m} = \mu g$.
2. **Angular Acceleration:** It creates a torque about the center of mass, $	au = f_k R = \mu mg R$, which starts spinning the disc. The angular acceleration is $\alpha = \frac{	au}{I} = \frac{\mu mg R}{\frac{1}{2}mR^2} = \frac{2\mu g}{R}$.

### The Master Equations

We can now write the kinematic equations for the disc at any time $t$ before pure rolling begins:
- Linear velocity: $v(t) = v_0 - \mu g t$
- Angular velocity: $\omega(t) = 0 + \left(\frac{2\mu g}{R}ight)t$

The magic happens at time $t = t_0$, when the slipping completely stops and pure rolling begins. The condition for pure rolling is that the velocity of the bottommost point must be zero, which mathematically translates to $v(t_0) = R \omega(t_0)$.

Equating our expressions:
$$v_0 - \mu g t_0 = R \left(\frac{2\mu g t_0}{R}ight)$$
$$v_0 - \mu g t_0 = 2\mu g t_0$$
$$3\mu g t_0 = v_0 \implies t_0 = \frac{v_0}{3\mu g}$$

Substituting this time back into our velocity equation gives the final velocity of the center of mass:
$$v = v_0 - \mu g \left(\frac{v_0}{3\mu g}ight) = \frac{2}{3}v_0$$

### Calculating the Work Done

Now for the second part: calculating the work done by friction. Instead of integrating force over the complex path of the contact point, we use the elegant **Work-Energy Theorem**. The work done by friction is simply the change in the total kinetic energy of the disc.

$$W_f = \Delta K = \left(\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2ight) - \frac{1}{2}mv_0^2$$

Substituting our time-dependent expressions for $v$ and $\omega$:
$$W_f = \frac{1}{2}m(v_0 - \mu g t)^2 + \frac{1}{2}\left(\frac{1}{2}mR^2ight)\left(\frac{2\mu g t}{R}ight)^2 - \frac{1}{2}mv_0^2$$

Expanding the brackets and collecting the terms, the initial kinetic energy $\frac{1}{2}mv_0^2$ cancels out beautifully. We are left with a neat quadratic expression in terms of time $t$:
$$W_f = \frac{1}{2}m\mu^2 g^2 t^2 - mv_0\mu g t + m\mu^2 g^2 t^2$$
$$W_f = \frac{3}{2}m\mu^2 g^2 t^2 - mv_0\mu g t = \frac{m\mu g t}{2}(3\mu g t - 2v_0)$$

### Final Calculation

Finally, what happens for a time much longer than $t_0$? Once pure rolling starts, the contact point is instantaneously at rest relative to the ground, so static friction does zero work! The total work done by friction is just the work done up to $t_0$.

Substituting $t_0 = \frac{v_0}{3\mu g}$ into our work expression:
$$W_{total} = \frac{m\mu g}{2}\left(\frac{v_0}{3\mu g}ight)\left(3\mu g \frac{v_0}{3\mu g} - 2v_0ight)$$
$$W_{total} = \frac{mv_0}{6}(v_0 - 2v_0) = -\frac{mv_0^2}{6}$$

This negative work represents the mechanical energy dissipated as heat during the slipping phase.

Visualized Solution

Initial State Analysis

Equations of Motion

Kinematics as a Function of Time

Condition for Pure Rolling

Time and Velocity for Pure Rolling

Work Done by Friction ( $t \leq t_{0}$ )

Substituting Kinematic Variables

Simplifying the Work Expression

Total Work Done ( $t ≫ t_{0}$ )

The Sigma Insight: Rolling Motion

Analyzing the Setup

The Master Equations

Calculating the Work Done

Final Calculation

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