Posted by ken on .
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,
(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) 
P.E 
Anonymous,
tennis
scoring.
zero/nothing iz called ? 
physics 
drwls,
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,
(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)