Question

A solid circular disc (m, R) translates with V_o and rotates with w_o = 2V₀/R at t = 0. Because of friction, pure rolling starts at t = t_o. The coefficient of friction is \(\mu\) find t₀

Answer 1

Correct option is (A) \(\frac{v_0}{ 3\mu g}\)

There is a friction force between the disc and the ground. Friction opposes the direction of motion. Since the bottom of the disc is moving to the left relative to the ground, that means friction force points to the right.

Take counterclockwise torques to be positive, and clockwise torques to be negative. Sum of the torques on the disc:

∑τ = Iα

-FR = \(\frac12\) mR² α

α = \(\frac{-2F}{mR}\)

α = \(\frac{-2mg\mu}{mR}\)

α = \(\frac{-2g\mu}{R}\)

The disc decelerates counterclockwise.  After time t₀, the angular velocity changes from ω₀ = \(\frac{2v_0}{R}\) to ω = \(\frac vR\).  Therefore:

ω = αt + ω₀

\(\frac vR\) = (\(\frac{-2g\mu}{R}\))t₀ + \(\frac{2v_0}{R}\)

v = -2gμ t₀ + 2v₀

Sum of the forces on the disc:

∑F = ma

F = ma

mgμ = ma

gμ = a

After time t₀, the disc accelerates from v₀ to v:

v = at + v₀

v = (gμ) t₀ + v₀

Two equations, two variables. Substitute this expression for v into our first equation:

gμ t₀ + v₀ = -2gμ t₀ + 2v₀

3gμ t₀ = v₀

t₀ = \(\frac{v_0}{ 3g\mu}\)