Posted by **ken** on Sunday, November 4, 2007 at 7:09pm.

A ball is attached to a string with length of L. It swings in a horizontal circle, with a constant speed. The string makes an angle (theta) with the vertical, and T is the magnitude of the tension in the string.

1)Determine the Mass of the Ball.

2)Determine th Speed of the Ball.

3)Determine the Frequency of revolutions of the Ball.

1)

F=Tsin(theta)=ma

m= (Tcos theta)/g

Tcos(theta)- mg= 0

T= mg/[cos(theta)]

F=Tsin(theta)

F= mg/[cos(theta)] * sin(theta) subtitute for T

F=mgtan(theta)

F=(mv^2)/r r=Lsin(theta)

mgtan(theta)=(mv^2)/ Lsin(theta)

v= square root of gLtan(theta)sin(theta)

3)

T= 2pi time square root of (l /g)

frequency= 1/t

frequency= 1/ [ 2pi time square root of (l /g) ]

I need comment about my solution. Thanks

- physics -
**drwls**, Sunday, November 4, 2007 at 7:44pm
(1) T cos theta = mg

m = (T/g) cos theta

(2) T sin theta = m V^2/r

= (T/g) cos theta V^2/r

tan theta = V^2/(gr)

V = sqrt [g r tan theta)

3) V * period = V/f = 2 pi r

f = [1/(2 pi)]* sqrt (g r tan theta)

- physics -
**drwls**, Monday, November 5, 2007 at 2:54am
Anonymous, you should have posted this as a new question, not under someone else's physics question

In tennis, zero is "love"

The expression "love" originated in the French word "l'ouef" (the goose egg), but in French tournaments today, the official uses the word "zero" when announcing the score.

- physics -
**annon**, Wednesday, April 20, 2011 at 8:06pm
(1) T cos theta = mg

m = (T/g) cos theta

(2) T sin theta = m V^2/r

= (T/g) cos theta V^2/r

tan theta = V^2/(gr)

V = sqrt [g r tan theta)

3) V * period = V/f = 2 pi r

f = [1/(2 pi)]* sqrt (g r tan theta)

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