Question

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∘?

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

Answer 2

The value of ω is 19.4 rad/s.

The normal reaction force N acts perpendicular to the circular wire, always directed towards the center.

Under equilibrium:

Ncosθ = mg                      ..........(1)

and Nsinθ = m$w^{2}$rsinθ    ...........(2)

dividing equation 1 by equation 2:

ω = sqrt{ g/rcosθ }

when θ = 30°

ω = sqrt{ g/rcos30 }

ω = sqrt{ 9.8/0.03 × cos30 }

ω = 19.4 rad/s

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