Question

A ring of mass $m$ is thrown in horizontal direction on a rough horizontal surface with a speed $v_0$. The speed of the ring, when it start pure rolling, is -

• A. $\frac{v_0}{2}$
• B. $\frac{v_0}{3}$
• C. $\frac{2v_0}{3}$
• D. $\frac{3v_0}{3}$

Answer 1

Answer: (A)

$\frac{v_0}{2}$

Answer 2

The problem describes a ring of mass $m$ thrown horizontally on a rough surface with an initial speed $v_0$. Initially, the ring has linear velocity but no angular velocity ($\omega_0 = 0$). Due to the rough surface, kinetic friction acts on the ring. This friction force will:

1. Oppose the linear motion: It will act in the direction opposite to $v_0$, causing the linear speed to decrease.

2. Cause rotation: It will exert a torque about the center of mass, causing the ring to start rotating.

Let $R$ be the radius of the ring.
Let $v_f$ be the final linear speed and $\omega_f$ be the final angular speed when the ring starts pure rolling.
For pure rolling, the condition is $v_f = R\omega_f$.

Let's use the impulse-momentum theorem for linear and angular motion.
The kinetic friction force is $f_k = \mu_k N$, where $N=mg$ (normal force) and $\mu_k$ is the coefficient of kinetic friction. So, $f_k = \mu_k mg$.

1. Linear Motion:
The friction force acts opposite to the initial velocity $v_0$.
According to the linear impulse-momentum theorem:
$\text{Impulse} = \Delta \text{Linear Momentum}$
$-f_k \Delta t = m(v_f - v_0)$
$f_k \Delta t = m(v_0 - v_f)$ (Equation 1)

2. Rotational Motion:
The friction force creates a torque about the center of mass. The torque $\tau = f_k R$. This torque causes an angular acceleration.
According to the angular impulse-momentum theorem:
$\text{Angular Impulse} = \Delta \text{Angular Momentum}$
$\tau \Delta t = I(\omega_f - \omega_0)$
For a ring, the moment of inertia about its center of mass is $I = mR^2$.
The initial angular velocity $\omega_0 = 0$.
So, $(f_k R) \Delta t = mR^2 (\omega_f - 0)$
$f_k R \Delta t = mR^2 \omega_f$
$f_k \Delta t = mR \omega_f$ (Equation 2)

3. Equating and Solving:
From Equation 1, we have $f_k \Delta t = m(v_0 - v_f)$.
Substitute this into Equation 2:
$m(v_0 - v_f) = mR \omega_f$
$v_0 - v_f = R \omega_f$

Now, apply the condition for pure rolling, $v_f = R\omega_f$.
Substitute $R\omega_f = v_f$ into the equation:
$v_0 - v_f = v_f$
$v_0 = 2v_f$
$v_f = \frac{v_0}{2}$

Alternatively, we can use a general formula for such problems. For an object with moment of inertia $I = k m R^2$ (where $k=1$ for a ring, $k=1/2$ for a disc/cylinder, $k=2/5$ for a solid sphere), thrown with initial linear velocity $v_0$ and zero initial angular velocity, the final velocity $v_f$ when pure rolling starts is given by: $v_f = \frac{v_0}{1+k}$ For a ring, $k=1$. $v_f = \frac{v_0}{1+1} = \frac{v_0}{2}$

The final speed of the ring when it starts pure rolling is $\frac{v_0}{2}$.