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

Answer 1

A. Tb = m(Vb/R^2 + g)

B. Tt = m(Vt/R^2 - g)

C. Difference is 2mg

Step-by-step explanation:

Detailed explanation and calculation is shown in the image below

Answer 2

a) Tb = (m*v b²/R) + (m*g)

b) Tt = (m*vt²/R) - (m*g)

c) Δ T = 6*m*g

Step-by-step explanation:

When an object undergoing uniform circular motion with uniform speed, It has an acceleration towards the center of the circle. This acceleration is known as the centripetal acceleration. It's magnitude equal to v²/r.

According to Newton's Second Law, the centripetal force acting on the object towards the center of the circle can be expressed as such:

Fc = m*ac = m*v²/r

a) Identify the given information in the problem:

• Mass of the balls, which is whirled in a vertical circle, is m

• The gravitational acceleration near the earth is g

• The radius of the circle is R

• The velocity of the ball at the bottom end of the vertical circle is v b

• The velocity of the ball at the top end of the vertical circle is v t

From the pic shown,

Part (a):

The net force acting on the ball whirling at the bottom of the circle:

Fnet_bottom = Tb - m*g

Where T b is the tension produced in the string when the ball is at the bottom end of the circle.

Since this net force supplies the centripetal force acting on the ball at the bottom of the circle. Therefore,

m*v b²/R = Tb - m*g

Simplifying it further, we will get:

Tb = (m*v b²/R) + (m*g)

then

Part (b):

The net force acting on the ball whirling at the top of the circle:

Fnet_top = - Tt - m*g

Where T t is the tension produced in the string when the ball is at the top of the circle.

Since this net force supplies the centripetal force acting on the ball at the top of the circle. Therefore,

- m*vt²/R = - Tt - m*g

Simplifying it further, we will get:

Tt = (m*vt²/R) - (m*g)

Part (c):

The tension produced in the string at the bottom will be greater than the tension produced in the string at the top by Δ T = T b − T t .

Thus,

Δ T = T b − T t

After substituting Tb and Tt, we have

Δ T = ((m*v b²/R) + (m*g)) - ((m*vt²/R) - (m*g))

⇒ Δ T = (m/R)*(v b² - vt²) + 2*m*g (i)

Now, by using the Conservation of Energy,

Kb + Ub = Kt + Ut

Where:

• Kb and Ub are the kinetic and the potential energies of the ball at the bottom of the circle respectively.

• Kt and Ut are the kinetic and the potential energies of the ball at the top of the circle respectively.

Thus,

0.5*m*vb² + m*g*(0) = 0.5*m*vt² + m*g*(2R)

⇒ (vb² - vt²) = 4*g*R

After substituting (vb² - vt²) into relation (i), we have:

Δ T = (m/R)*(4*g*R) + 2*m*g = 4*m*g + 2*m*g

Simplifying it further, we will get:

Δ T = 6*m*g