A tennis ball connected to a string is spun around in a vertical, circular path

Question

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 = 0.174 kg and moves at v = 5.16 m/s. The circular path has a radius of R = 1.13 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?

Explanation / Answer

here,

mass , m = 0.174 kg

v = 5.16 m/s

radius , R = 1.13 m

a)

the magnitude of the tension in the string when the ball is at the bottom of the circle , Tb = m * v^2 /R + m * g

Tb = 0.174 * ( 5.16^2/1.13 + 9.81) N

Tb = 5.8 N

b)

the magnitude of the tension in the string when the ball is at the side of the circle , T = sqrt((m * v^2 /R)^2 + (m * g)^2)

Ts = sqrt( 0.174^2 * ( (5.16^2/1.13)^2 + (9.81)^2))

Ts = 4.44 N

c)

the magnitude of the tension in the string when the ball is at the top of the circle , Tt = m * v^2 /R - m * g

Tt = 0.174 * ( 5.16^2/1.13 - 9.81) N

Tt = 2.4 N

d)

let the minimum speed be v

for the string to not to slack

m * v^2 /R = m * g

v = sqrt(R*g)

v = sqrt(1.13 * 9.81) = 3.33 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.