# Homework Help: Angular Speed of a bob of mass

1. Dec 8, 2014

### David Swift

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.)

3. 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?

2. Dec 8, 2014

### Staff: Mentor

Almost. Realize that the centripetal acceleration is horizontal, so only the horizontal component of the tension provides the centripetal force.

3. Dec 8, 2014

### David Swift

I am not sure what you mean by that. Have I used a wrong equation?

4. Dec 8, 2014

### Staff: Mentor

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.

5. Dec 8, 2014

### David Swift

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

6. Dec 8, 2014

### haruspex

You seem to have taken a step backwards. Your original equation for tension(F) was correct:
Doc Al is telling you this equation is wrong (assuming F is still the tension):
Please try to post a correct version.