The 5-kg cylinder is initially at rest when it is placed in contact with the wall b and the rotor at a. if the rotor always maintains a constant clockwise angular velocity v = 6 rad> s, determine the initial angular acceleration of the cylinder. the coefficient of kinetic friction at the contacting surfaces b and c is mk = 0.2.

Explanation:

You forgot to include the diagram, but I'm assuming it's the one attached to this answer.

First, sum of the torques equals moment of inertia times angular acceleration.

∑τ = Iα

Fa r - Fb r = (1/2 mr²) α

Fa - Fb = 1/2 m r α

Na μ - Nb μ = 1/2 m r α

μ (Na - Nb) = 1/2 m r α

α = 2μ (Na - Nb) / (m r)

To find the normal forces, we need to take sum of the forces in the x and y directions.

∑Fₓ = maₓ

Nb - Na sin 45 + Fa cos 45 = 0

Nb - Na sin 45 + Na μ cos 45 = 0

Nb - Na (√2 / 2) + Na μ (√2 / 2) = 0

Nb - Na (√2 / 2) (1 + μ) = 0

Nb = Na (√2 / 2) (1 + μ)

∑Fᵧ = maᵧ

Fb - W + Na cos 45 + Fa sin 45 = 0

Nb μ - W + Na cos 45 + Na μ sin 45 = 0

Nb μ - W + Na (√2 / 2) + Na μ (√2 / 2) = 0

Nb μ - W + Na (√2 / 2) (1 + μ) = 0

Substituting:

Nb μ - W + Nb = 0

Nb (μ + 1) = W

Nb = mg / (μ + 1)

Nb = (5 kg) (9.8 m/s²) / (0.2 + 1)

Nb = 40.83 N

Finding Na:

mg / (μ + 1) = Na (√2 / 2) (1 + μ)

Na = mg√2 / (μ + 1)²

Na = (5 kg) (9.8 m/s²) √2 / (0.2 + 1)²

Na = 48.12 N

Therefore:

α = 2μ (Na - Nb) / (m r)

α = 2(0.2) (48.12 - 40.83) / (5 × 0.125)

α = 4.67 rad/s²

Rounding to 1 sig-fig, the angular acceleration is 5 rad/s².

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