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, and R for radius of the circle.)

(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.)

(c) How much greater is the tension at the bottom? (Use the following as necessary: m, g.)

To solve this problem, we can use the principle of conservation of energy, which states that the total energy of a system remains constant.

(a) To find the tension in the string at the bottom, we first need to determine the ball's kinetic and potential energy at this point. At the bottom of the vertical circle, the ball's velocity (vb) is at its maximum, and its height above the center of the circle is the radius (R).

The kinetic energy (KE) of the ball at the bottom is given by KE = (1/2)mvb^2, where m is the mass of the ball.

The potential energy (PE) of the ball at the bottom is given by PE = mgh, where g is the gravitational acceleration and h is the height of the ball above the reference point (in this case, the center of the circle). Since the height at the bottom is equal to the radius (R), we have PE = mgh = m g R.

According to the conservation of energy, the total energy (TE) of the system is constant and equal to the sum of the kinetic and potential energy: TE = KE + PE. Since TE remains constant, we have TE = KE = PE.

Therefore, at the bottom:
TE = (1/2)mvb^2 + m g R.

Since TE = KE, we can solve for vb:
vb^2 = 2gR.

The tension in the string at the bottom (Tb) is equal to the centripetal force required to keep the ball in circular motion. Using Newton's second law (F = m*a), where a is the centripetal acceleration, we have Tb - m*g = m*a.

At the bottom, the centripetal acceleration (ab) is given by ab = vb^2/R. Plugging in the values, we have:

Tb - m*g = (m*vb^2)/R.

Now, we substitute vb^2 = 2gR:

Tb - m*g = (m*2gR)/R.

Simplifying the equation gives:

Tb - m*g = 2mg.

Therefore, the tension in the string at the bottom is:

Tb = 3mg.

(b) To find the tension in the string at the top, we use the same approach. At the top of the vertical circle, the ball's velocity (vt) is at its lowest, which means its kinetic energy is also at its lowest. The potential energy at the top is the maximum, as the ball is at the maximum height above the reference point.

The kinetic energy (KE) at the top is given by KE = (1/2)mvb^2, where vb is the velocity at the bottom.

The potential energy (PE) at the top is given by PE = mgh, where h is the height of the ball above the reference point. At the top of the vertical circle, the height (h) is equal to 2R (twice the radius).

Using the conservation of energy (TE = KE + PE), we have TE = (1/2)mvb^2 + mgh = KE + PE. Since TE remains constant, we have TE = KE = PE.

Therefore, at the top:
TE = (1/2)mvb^2 + mgh = mgh + mgh = 2mgh.

The tension in the string at the top (Tt) is equal to the centripetal force required to keep the ball in circular motion. Using Newton's second law (F = m*a), where a is the centripetal acceleration, we have Tt + m*g = m*a.

At the top, the centripetal acceleration (at) is given by at = vt^2/R. Plugging in the values, we have:

Tt + m*g = (m*vt^2)/R.

Now, we substitute vt^2 = 2gR:

Tt + m*g = (m*2gR)/R.

Simplifying the equation gives:

Tt + m*g = 2mg.

Therefore, the tension in the string at the top is:

Tt = mg.

(c) To calculate the difference in tension between the bottom and the top, we subtract the tension at the top (Tt) from the tension at the bottom (Tb):

Difference in tension = Tb - Tt = 3mg - mg = 2mg.

Therefore, the tension at the bottom is twice as great as the tension at the top.