Question

A ball of mass m = 0.1kg is connected to a rope of length L = 1.2 m. The ball is swung around in a vertical circle and ball is moving around the circle at a constant speed v (uniform circular motion). The tension on the rope at the top of the string is Ttop = 2 mg = 1.96 N, where g = 9.8 m/sec2 is the acceleration due to gravity.
Problem 1: What is v, the speed of the ball?
What is Tbottom, the tension on the rope at the bottom of the circle?

Answer 1

Answer:

The speed of the ball is approximately 5.94 m/s

The Tension of the string at the bottom is 3.92 N

Explanation:

We need to find the speed of the ball, which is constant due to the fact that we are in a uniform circular motion. Notice as well that the speed of the ball is the magnitude of the tangential velocity "$v_t$" (vector that changes direction with the position of the ball but doesn't change magnitude in this case).

We analyze first the top position of the circular motion, for which information on the tension of the string is given (see first free body diagram in the attached picture).  We are told that the tension at the top of the movement equals twice the force of gravity on the ball's mass: T - 2*m*g = 1.96 N. And we know that there are two forces acting on the ball in that position (illustrated with the green arrows pointing down): one is the ball's weight due to gravity, and the other is the string's tension. So we can write Newton's second law for this situation:

$F_{net}= T_{top}+W\\F_{net}=2\,W+W\\F_{net}=3\,W\\F_{net}=2.94 N$

Newton's second law tells us that the net force should equal the mass of the ball times its acceleration (F = m * a), and in this motion, the acceleration is the centripetal acceleration. Therefore weuse this equation to solve for the centripetal acceleration of the ball:

$m\,a_c=2.94\,N\\a_c=\frac{2.94\,N}{0.1\,kg} \\a_c=29.4\,\frac{m}{s^2}$

The centripetal acceleration is defined as the square of the tangential velocity divided the radius of the circular motion. Then we use it to derive the magnitude of the tangential velocity (speed of the ball):

$a_c=\frac{v^2}{R} \\29.4\,\frac{m}{s^2} =\frac{v_t^2}{R} \\v_t^2=29.4\,(1.2)\,\frac{m^2}{s^2} \\v_t=5.94\,\frac{m}{s}$

So we have found the speed of the ball.

Now we focus our attention to the bottom of the motion, and again use Newton's second law to solve for the string tension (see second free body diagram in the attached picture).

We notice here that the tension and the weight are acting in opposite directions, so we have such into account when finding the net force on the ball, and then solve for the tension knowing the value of the centripetal acceleration (recall that the magnitude of the tangential velocity is the same because of the uniform circular motion).

$F_{net}= T_{bot}-W\\m\,a_c=T_{bot}-0.98\,N\\2.94\,N=T_{bot}-0.98\,N\\T_{bot}=(2.94+0.98)N\\T_{bot}=3.92\,N$