Question

A disc of radius R is rotating with an angular speed ωo about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is µ_k.

1. What was the velocity of its centre of mass before being brought in contact with the table?
2. What happens to the linear velocity of a point on its rim when placed in contact with the table?
3. What happens to the linear speed of the centre of mass when disc is placed in contact with the table?
4. Which force is responsible for the effects in (b) and (c).
5. What condition should be satisfied for rolling to begin?
6. Calculate the time taken for the rolling to begin.

Answer 1

a. Before being brought in contact with the table the disc was in pure rotational motion hence, v_CM = 0.

b. When the disc is placed in contact with the table due to friction velocity of a point on the rim decreases.

c. When the rotating disc is placed in contact with the table due to friction centre of mass acquires some linear velocity.

d. Friction is responsible for the effects in (b) and (c) .

e. When rolling starts v_CM = ωR.

Where ω is the angular speed of the disc when rolling just starts.

f. Acceleration produced in the centre of mass due to friction

a_CM = ${\displaystyle \frac{F}{M}-\frac{{\mu }_k mg}{m}={\mu }_{k}g}$

Angular retardation is produced by the torque due to friction.

α = ${\displaystyle \frac{\tau }{I}=\frac{{\mu }_k mgR}{I}}$  ......[∵ τ = (μ_kN)R = μ_kmgR]

∴ v_CM = u_CM + a_CMt

⇒ v_CM =  μ_kgt  ......(∵ u_CM = 0)

And ω = ω₀ + αt

⇒ ${\displaystyle \omega ={\omega }_0-\frac{{\mu }_{k}mgR}{I}t}$

For rolling without slipping, ${\displaystyle \frac{v_{CM}}{R}}$ = ω

 ⇒ ${\displaystyle \frac{v_{CM}}{R}={\omega }_0-\frac{{\mu }_{k}mgR}{I}t}$

${\displaystyle \frac{{\mu }_{k}gt}{R}={\omega }_0-\frac{{\mu }_{k}mgR}{I}t}$

t = ${\displaystyle \frac{R{\omega }_0}{{\mu }_{k}g(1+\frac{m{R}^2}{I})}}$