Question

The
maximum tension that a 0 50 m
string can tolerate is 14 N A 0 25 kg
ball attached to this string is being
whirled in a vertical circle What is the
maximum speed the ball can have (a) at
the top of the circle. (b)at the bottom
of the circle?

Answer 1

a) At the top of the circle, the tension in the string will be equal to the weight of the ball (mg). So we can set up an equation:

Tension = mg = (0.25 kg)(9.8 m/s^2) = 2.45 N

The maximum speed the ball can have is when the tension in the string is equal to the maximum tension it can tolerate, so we can set up another equation:

Tension = Maximum tension = 14 N

Using these two equations, we can solve for the maximum speed:

Tension = mv^2/r

14 N = (0.25 kg) v^2 / 0.50 m

v^2 = (14 N)(0.50 m) / 0.25 kg

v^2 = 28 m^2/s^2

v = √(28 m^2/s^2) = 5.3 m/s

Therefore, the maximum speed the ball can have at the top of the circle is 5.3 m/s.

b) At the bottom of the circle, the tension in the string will be equal to the weight of the ball plus the centripetal force needed to keep it moving in a circle (mg + mv^2/r). So we can set up an equation:

Tension = mg + mv^2/r = (0.25 kg)(9.8 m/s^2) + (0.25 kg) v^2 / 0.50 m

The maximum speed the ball can have is again when the tension in the string is equal to the maximum tension it can tolerate, so we can set up another equation:

Tension = Maximum tension = 14 N

Using these two equations, we can solve for the maximum speed:

14 N = (0.25 kg)(9.8 m/s^2) + (0.25 kg) v^2 / 0.50 m

14 N - (0.25 kg)(9.8 m/s^2) = (0.25 kg) v^2 / 0.50 m

v^2 = [2(14 N - (0.25 kg)(9.8 m/s^2))] / 0.25 kg

v^2 = 46.84 m^2/s^2

v = √(46.84 m^2/s^2) = 6.8 m/s

Therefore, the maximum speed the ball can have at the bottom of the circle is 6.8 m/s.