What is the Angular Speed of the Bob?

AI Thread Summary
The discussion centers on calculating the angular speed of a bob performing uniform circular motion. The bob has a mass of 0.5 kg and is attached to a string of length 0.5 m, making a 30° angle with the vertical. Initial calculations for tension in the string yield 5.66 N, and the radius of the circular motion is determined to be 0.25 m. However, it is clarified that only the horizontal component of the tension contributes to the centripetal force, leading to a correction in the angular speed calculation. The final angular speed is suggested to be 2.22 rad s^-1 after addressing the correct application of forces.
David Swift
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1. Relevant information

A bob of mass 0.5kg is attached to one end of a light inextensible string of length 0.500 m, whose other end is attached to a fixed pivot. The bob performs uniform circular motion in a horizontal plane, with the string making an angle of 30.0° with the vertical.

2. Question

What is the angular speed of this circular motion?
Hint: you can begin by using Newton's second law in the vertical direction to find the tension in the string.)

The Attempt at a Solution



Tension in string
F Cos 30 = 0.5 x 9.81
F = (0.5 x 9.81) / Cos 30
F = 5.66N

Radius of circle must be 0.5x Sin 30 = 0.25m

F = mass x radius x angular speed^2
Angular speed = Sqrt ( 5.66 / (0.5 x 0.25))
Angular speed = 6.72 rad s^-1

Have I done this right?
 
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David Swift said:
Tension in string
F Cos 30 = 0.5 x 9.81
F = (0.5 x 9.81) / Cos 30
F = 5.66N

Radius of circle must be 0.5x Sin 30 = 0.25m

F = mass x radius x angular speed^2
Angular speed = Sqrt ( 5.66 / (0.5 x 0.25))
Angular speed = 6.72 rad s^-1

Have I done this right?
Almost. Realize that the centripetal acceleration is horizontal, so only the horizontal component of the tension provides the centripetal force.
 
Doc Al said:
Almost. Realize that the centripetal acceleration is horizontal, so only the horizontal component of the tension provides the centripetal force.
I am not sure what you mean by that. Have I used a wrong equation?
 
David Swift said:
I am not sure what you mean by that. Have I used a wrong equation?
When you used ##F = m \omega^2 r##, you used the entire tension as the force. But only the horizontal component of the tension creates the centripetal acceleration.
 
Doc Al said:
When you used ##F = m \omega^2 r##, you used the entire tension as the force. But only the horizontal component of the tension creates the centripetal acceleration.
F(x) = ((m x g ) x Xcomponent / length
F (x) = (0.5 x 9.81 x 0.5 sin 30 ) / 0.5m = 2.22 rad s^-1
 
David Swift said:
F(x) = ((m x g ) x Xcomponent / length
You seem to have taken a step backwards. Your original equation for tension(F) was correct:
David Swift said:
F Cos 30 = 0.5 x 9.81
Doc Al is telling you this equation is wrong (assuming F is still the tension):
David Swift said:
F = mass x radius x angular speed^2
Please try to post a correct version.
 
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