Chapter 8: Q. 48 (page 201)

The 10 mg bead in FIGURE P8.48 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
What is ω_c in rpm for the loop shown in the figure?
At what value of ω_c in rpm is θ=30^o

Short Answer

Expert verified

1) Angular Speed of the ball is $ω_{c} = \sqrt{\frac{g}{R \cos θ}}$

2) The angular speed ω_c in rpm is 143.52 rpm.

Step by step solution

Part(1) Step 1 : Given Information

R = 5cm =0.05m

θ=30^o

Part(1) Step 2: Explanation

First draw the Free body diagram and equate forces components.

$N \cos θ = m g . . . . . . . . . . . . . (1)$

and

$N s i n θ = m ω^{2} r . . . . . . . . . . . . . . . . . . . (2)$

Divide (2) by (1) we get

$t a n θ = \frac{ω^{2} r}{g} S u b s t i t u t e r = R \sin θ t a n θ = \frac{ω^{2} R \sin θ}{g}$

Solve it for ω, we get

$ω = \sqrt{\frac{g}{R \cos θ}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)$

Part(2) Step 1 : Given Information

R = 5cm=0.05m

θ=30^o

Part(2) Step 2 : Explanation

Substitute the values in the equation (2), we get

$ω = \sqrt{\frac{(9.8 m / s^{2})}{(0.05 m \times \cos 30^{0})}} = 15.03 r a d / s e c$

Convert in rpm

$ω = (15.03 r a d / s e c) (\frac{1 r p m}{2 π r a d}) (\frac{60 s e c}{1 m i n}) = 143.52 r p m$

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Short Answer

Step by step solution

Part(1) Step 1 : Given Information

Part(1) Step 2: Explanation

Part(2) Step 1 : Given Information

Part(2) Step 2 : Explanation

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