A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.175 kg and moves at v = 5.22 m/s. The circular path has a radius of R = 1.14 m
1.What is the magnitude of the tension in the string when the ball is at the bottom of the circle?
2.What is the magnitude of the tension in the string when the ball is at the side of the circle?
3.What is the magnitude of the tension in the string when the ball is at the top of the circle?
4.What is the minimum velocity so the string will not go slack as the ball moves around the circle?
Centripetal Force (C) = mv^2/r = (0.175)*(5.22)^2 * (1/1.14) =
4.18286N
Gravitational force (mg) = 0.175*9.81 = 1.71675 N
1. ball is at the bottom : Tension = C+mg = 5.899N
2. ball is at the side : Tension = C = 4.182866N
3 . At the top : Tension = C-mg = 2.46611 N
minimum velocity : v^2 / r = g or v = sqrt(rg) = sqrt ( 1.14*9.81)
= 3.3441m/s
a) T - mg = M*v*v/R
T = mg + m*v*v/R = 5.9 N
b) T = m*v*v/R = 4.18 N
c) T + mg = m*v*v/R
T = m*v*v/R - mg = 4.18 - 0.175*9.8 = 2.47 N
d) The chances of string getting slacked is at top point.
For minimum velocity, T = 0.
v = sqrt(Rg) = 3.342 m/s
SOLUTION :
1.
When the ball is at the bottom
T = mg + mv^2/r = m(g + v^2/r) = 0.175(9.8 + 5.22^2 / 1.14) = 5.90 N (ANSWER)
2.
When the ball is at the side of the circle :
T = mv^2/r = 0.175(5.22^2 / 1.14) = 4.18 N (ANSWER)
3.
When the ball is at the top of the circle :
T = mv^2/r - mg = 0.175(5.22^2 / 1.14) - 0.175*9.80= 2.47 N (ANSWER)
4.
At the top of circle for a given v, T is the least.
So,
When T = 0 when ball is at the top, will be the point after which string gets slack..
So,
T = 0
=> mv^2/r - mg = 0
=> v^2 = g r
=> v = sqrt(g r)
=> v = sqrt(9.80 * 1.14) = 3.34 m/s
So, minimum velocity is 3.34 m/s so the string will not go slack. (ANSWER).
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