Question

A 1-kg ball is tied to the end of a 2-meter string and revolved in a horizontal plane making a 30^o angle with the vertical.

(a) What is the ball's speed?

(b) If the ball is now revolved so that its speed is 4m/s, what angle does the string need to make with the vertical?

(c) If the string can withstand a maximum tension of 10 N, what is the highest speed at which the ball can travel?

Answer 1

The trajectory of the ball is an horizontal circle of radius R = L sin() where L (= 2 m) is the length of the string and is the angle which it makes with the vertical.

If the ball travels at speed v, it undergoes a centripetal (horizontal) acceleration v^2/R where R is the radius of curvature of the trajectory-- which says that the above formula always give the component of the acceleration which is perpendicular to the trajectory, even when the speed is not constant and/or when the trajectory is not a circle.This is the basis for the following answers to your questions:

(a) If g is the acceleration of gravity then v^2/R = g tan().
So, v^2 = Rg tan() = gL sin() tan()
With = 30° and g = 9.80665 m/s^2 , a length L=2 m gives

v = 9.8*2*sin30 tan30=5.66 m/s

(b) The formula is v^2 = gL sin() tan() = gL (1-x^2)/x where x is cos().

This is a quadratic equation in x, namely:
x^2 + (v^2/gL) x -1 = 0
Letting v be 4 m/s and solving for a positive value of x below 1, we obtain x = 0.67 and, therefore, A = 4.79° (almost exactly).

(c) The tension of the rope is mg / cos() = mg / x.
When m is 1 kg, this is equal to 10 N when x is about .98

so = cos^{-1}x=11.48

Using the before mentioned relation

v^2 = gL (1-x^2)/x, we obtain the value of v at which the rope would break, namely: v = .9 m/s.