Question

A solid cylinder having radius \(R\) and length \(L\) is slipping on a rough horizontal plane. At time \(t = 0\) the cylinder has a translational velocity \(v_0 = 49\) m/s, perpendicular to its axis and a rotational velocity \(v_0/4R\) about the centre. The time taken by the cylinder to start rolling is ________ seconds. (coefficient of kinetic friction \(\mu_k = 0.25\) and \(g = 9.8\) m/s²)

• A. 15
• B. 5
• C. 10
• D. 7.5

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The cylinder initially has both translational and rotational motion but is slipping ($v_0>R\omega_0$). Kinetic friction will act in the backward direction to reduce the translational velocity while simultaneously providing a torque that increases the angular velocity. Pure rolling begins when the condition $v(t) = R\omega(t)$ is met.
Step 2: Key Formula or Approach:
1. Linear motion: $v(t) = v_0 - at$, where $a = \mu_k g$.
2. Rotational motion: $\omega(t) = \omega_0 + \alpha t$, where torque $\tau = f_k R = I\alpha$.
3. Moment of inertia of a solid cylinder: $I = \frac{1}{2}mR^2$.
4. Pure rolling condition: $v(t) = R\omega(t)$.
Step 3: Detailed Explanation:
Given:
Initial velocity $v_0 = 49\text{ m/s}$.
Initial angular velocity $\omega_0 = \frac{v_0}{4R} = \frac{49}{4R}$.
Friction force $f_k = \mu_k mg = 0.25 m (9.8) = 2.45 m$.
Deceleration in translation $a = \frac{f_k}{m} = 2.45\text{ m/s}^2$.
The velocity as a function of time is:
$v(t) = v_0 - at = 49 - 2.45t$.
The torque provided by friction is $\tau = f_k \times R = (2.45m)R$.
Angular acceleration $\alpha = \frac{\tau}{I} = \frac{2.45 m R}{\frac{1}{2}mR^2} = \frac{4.9}{R}\text{ rad/s}^2$.
The angular velocity as a function of time is:
$\omega(t) = \omega_0 + \alpha t = \frac{49}{4R} + \frac{4.9}{R} t$.
For pure rolling to start, the velocity of the contact point must be zero, which means $v(t) = R\omega(t)$.
Substitute the equations into the condition:
$49 - 2.45t = R\left(\frac{49}{4R} + \frac{4.9}{R} t\right)$.
$49 - 2.45t = \frac{49}{4} + 4.9t$.
$49 - 2.45t = 12.25 + 4.9t$.
Group the terms with $t$ on one side and constants on the other:
$49 - 12.25 = 4.9t + 2.45t$.
$36.75 = 7.35t$.
Solve for $t$:
$t = \frac{36.75}{7.35} = 5\text{ s}$.
Step 4: Final Answer:
The time taken to start rolling is $5\text{ s}$.