Question

A ball whirls around in a vertical circle at the end of a string. The other end of the string is fixed at the center of the circle. Assume that the total energy of the ball-Earth system remains constant.

(a) What is the tension in the string at the bottom? (Use the following as necessary: m for mass of the ball, g for gravitational acceleration, vb for velocity at the bottom, andR for radius of the circle.)
Tb =____________.
(b) What is the tension in the string at the top? (Use the following as necessary: m, g, vt for velocity at the top, and R.)
Tt =______________.
(c) How much greater is the tension at the bottom? (Use the following as necessary: m, g.)
Tb = Tt +____________.

Answer 1

a) Tb = m (g + vb² / R) , b) Tt = m (g - vt² /R), c) Tb = Ty + (vb² -vt²) /R

Step-by-step explanation:

a) For this exercise we will use Newton's second law, at the bottom

Tb - W = m a

Where the acceleration is centripetal

a = v² / R

Tb = W + m vb² / r

Tb = m (g + vb² / R)

b) the tension at the top

-Tt -w = m a

Tt = W - m vt² / R

Tt = m (g - vt² / R)

c) write the two equations and solve system

Tb = m (g + vb² / R)

Tt = m (g - vt² / R) (-1)

We multiply by (1) and add

Tb -Ty = vb² /R -vt² /R

Tb = Ty + (vb² -vt²) / R