A 10 mg bead is free to slide on a frictionless wire loop. The loop rotates about a vertical axis with angular velocity ω. If ω is less than some critical value ωc, the bead sits at the bottom of the spinning loop. When ω>ωc, the bead moves out to some angle θ. Suppose R = 3.0 cm. What is ωc in rpm for the loop? At what value of ω, in rpm, is θ = 30∘?

Question

Amelie Brock

Physics

21 November, 20:03

A 10 mg bead is free to slide on a frictionless wire loop. The loop rotates about a vertical axis with angular velocity ω. If ω is less than some critical value ωc, the bead sits at the bottom of the spinning loop. When ω>ωc, the bead moves out to some angle θ. Suppose R = 3.0 cm. What is ωc in rpm for the loop? At what value of ω, in rpm, is θ = 30∘?

+1

Answers (1)

Know the Answer?

Maximus Kemp · Answer 1

Answer: 19.4 rad/sec

Explanation:

Weight of the bead: W=10mg

Radius of the circular loop is:

R = 3/100 = 0.03m

Angle = 30°

w=sqrt (mg/Rcos-tita

w=sqrt (9.8/0.03*cos (30°))

w=sqrt (377.2)

w=19.4 rad/sec