Ncert Exercises & Solutions

Q. 25 A disc of radius R is rotating with an angular $ω_{o}$ about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is $μ_{k}$ .

(a) What was the velocity of its center of mass before being brought in contact with the table?

(b) What happens to the linear velocity of a point on its rim when placed in contact with the table?

(d) Which force is responsible for the effects in (b) and (c)?

(e) What condition should be satisfied with rolling to begin?

(f) Calculate the time taken for the rolling to begin.

Hint: For pure rolling

v_{C M} = R ω .

Step 1: Find the velocity of center of mass Before being brought in contact with the table.

(a) Before being brought in contact with the table the disc was in pure rotational motion hence,

v_{CM} = 0

Step 2: When the disc is placed in contact.

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

Step 3: When the rotating disc is placed in contact with the table.

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

Step 4: Find which force is responsible for the effect in (b) and (c)

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

Step 4: write the equation for pure roling.

(e) When rolling starts

v_{C M} = ω R

where

ω

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

Step 5: Find the time taken for the rolling to begin.

(f) Acceleration produced in centre of mass due to friction

a_{CM} = \frac{F}{m} = \frac{μ_{k} mg}{m} = μ_{k} g

Angular retardation produced by the torque due to friction.

\begin{array}{l} α = \frac{τ}{I} = \frac{μ_{K} mgR}{I} [∵ τ = (μ_{k} N) R = μ_{k} mgR] \\ ∴ V_{CM} = u_{CM} + a_{CM} t \\ \Rightarrow V_{CM} = μ_{k} gt (∵ u_{CM} = 0) \\ and ω = ω_{0} + αt \\ \Rightarrow ω = ω_{0} - \frac{μ_{k} mgR}{I} t \end{array}

For rolling without slipping,

\frac{V_{CM}}{R} = ω

\begin{array}{l} \Rightarrow \frac{v_{CM}}{R} = ω_{0} - \frac{μ_{k} mgR}{I} t \\ \frac{μ_{k} gt}{R} = ω_{0} - \frac{μ_{k} mgR}{l} t \\ t = \frac{{Rω}_{0}}{μ_{k} g (1 + \frac{{mR}^{2}}{1})} \end{array}

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