- #1
Phoenixtears
- 82
- 0
SOLVED
A 530 g ball swings in a vertical circle at the end of a 2.9 m-long string. When the ball is at the bottom of the circle, the tension in the string is 21 N. What is the speed of the ball at that point?
2nd law statements
a = V^2/r
V= (circu.)/(period)
I began by drawing a force diagram as if the ball were at the bottom of the cycle. Therefor I would have Tension facing up and weight down. (The tension vector would be larger than the weight vector). There would be no normal because there is no force to act upon the ball in the normal direction. Then I concluded that the second law statements were as follows:
Max=0
May= T-mg
M(V^2/r)= T- mg (substitute in for a)
Then I just plugged in the numbers:
530(V^2/2.9)= 21- (530*9.8)
V= 5.32
However, this is not the answer. Where did I go wrong?
Thank you in advance!
~Phoenix
Homework Statement
A 530 g ball swings in a vertical circle at the end of a 2.9 m-long string. When the ball is at the bottom of the circle, the tension in the string is 21 N. What is the speed of the ball at that point?
Homework Equations
2nd law statements
a = V^2/r
V= (circu.)/(period)
The Attempt at a Solution
I began by drawing a force diagram as if the ball were at the bottom of the cycle. Therefor I would have Tension facing up and weight down. (The tension vector would be larger than the weight vector). There would be no normal because there is no force to act upon the ball in the normal direction. Then I concluded that the second law statements were as follows:
Max=0
May= T-mg
M(V^2/r)= T- mg (substitute in for a)
Then I just plugged in the numbers:
530(V^2/2.9)= 21- (530*9.8)
V= 5.32
However, this is not the answer. Where did I go wrong?
Thank you in advance!
~Phoenix
Last edited:
(And to think I've been changin to correct S.I. units all day...)