Circular motion of ball and string

[ScienceGeek24]

Homework Statement

Part A : A 0.5 Kg ball is attached to a cord 80 cm long. The ball is whirled in a vertical circle at a constant speed. If the ball takes 0.4 seconds to go around once, what is the tension in the cord at the top and bottom of the circle?

Part B How long does it take for the ball to go around once if the tension in the cord at the top of the circle is zero?

Homework Equations

v=2Pi*R/T
T=m(g+v^2/r) (bottom of the circle)
T=m(g-v^2/r) (top of the circle)

The Attempt at a Solution

For part A i used both of the circular motion formulas to fin the tension and i got totally different answers from what it suppose to be for top and bottom. Top it suppose to be 93.8N and bottom it suppose to be 104 N i got 92.7N top and 102.7 N bottom. ( I do not know if the answer is a print error).

For part A I tried to use v=2Pi*R/T to get the time and i keep getting the same time in part A which is 0.4 secs. But the answer is 1.8 seconds.

Can someone explain to be what I'm doing wrong? thanks!

 

[tony24810]

the formula for top of the circle should be

T=m(v^2/r - g) (top of the circle)

for part a, the time is 12.57 second

g = 9.81

try to put the numbers in your calculator more carefully, the result should come out correct

 

[tony24810]

for part b, tension must equal zero yet F=ma must be obeyed

F = m * a

T + mg = m * v^2/r (the downward force consist of tension and weight)

mg = m*v^2/r (T=0)

g = v^2/r

v = 2.8 m s^-1

now try the v=2Pi*R/T again

 

[ScienceGeek24]

Thanks it worked!