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 = \(\frac{v_0}{ 3g\mu}\)