A mass m rests on a turn-table at distance r = 0.5 m from the axis of rotation. The coefficient of static friction is mu_s = 0.32. Taking g = 9.8 m/s^2, what is the maximum angular speed omega before the mass slips?
Answer and explanation
Correct answer: D. 2.5 rad/s
- Step 1: Static friction provides the necessary centripetal force: f_s = m omega^2 r.
- Step 2: Maximum static friction: f_{s,max} = mu_s m g.
- Step 3: Set equal: m omega^2 r = mu_s m g, so omega^2 = mu_s * g / r.
- Step 4: Substitute values: omega^2 = (0.32 * 9.8) / 0.5 = 3.136 / 0.5 = 6.272.
- Step 5: omega = sqrt(6.272) = 2.504 rad/s.
Answer
2.5 rad/s.
Why not the others: 3.1 rad/s computes sqrt(mu_s g); 1.8 rad/s divides by 2r; 4.0 rad/s uses linear speed instead of angular speed.
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