A uniform cylinder of radius is spun about its axis to an angular velocity and then placed in a corner. The coefficient of friction between the wall (and floor) and the cylinder is . How many radians does it spin through before stopping?
Assumptions and Details
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Given:
→ μ = 3 → ω 0 = 2 π → R = 0 . 2 5 m → g = 1 0 m s − 2
Let m be the mass of the cylinder, F 1 be the frictional force acting along the floor and F 2 be the force acting along the wall as shown in the following figure. Equating components of force and normal reaction horizontally and vertically and using the laws of friction we get: N 1 + F 2 − m g = 0 ⋯ ( 1 ) F 1 − N 2 = 0 ⋯ ( 2 ) F 1 = μ N 1 ⋯ ( 3 ) F 2 = μ N 2 ⋯ ( 4 ) Using (1), (2), (3) and (4) we get: F 1 = 1 + μ 2 μ m g and F 2 = 1 + μ 2 μ 2 m g Moment of inertia, I is equal to I = 2 m R 2 Net torque acting on the cylinder τ n e t is ( α is the angular acceleration): τ n e t = − F 1 R − F 2 R = I α ⇒ I α = μ m g 1 + μ 2 1 + μ ⇒ α = R ( 1 + μ 2 ) − 2 μ g ( 1 + μ ) α and τ n e t are with a negative sign because the direction of ω 0 was assumed to be positive and the τ n e t and α acting are opposing it.
Now, when the circular motion of the cylinder stops, ω t 2 = 0 = ω 0 2 + 2 α θ ⇒ θ = 4 μ g ( 1 + μ ) ω 0 2 ( 1 + μ 2 ) R = 4 × 1 0 ( 3 + ( 3 ) 2 ) 4 π × ( 1 + ( 3 ) 2 ) × 0 . 2 5 ⇒ θ = 0 . 0 6 6 3 9 rad ≈ 0 . 0 6 6 5 rad