A 0.25 kg ball is suspended from a light 0.85 m string as shown. The string makes an angle of 28° with the vertical. Let U = 0 when the ball is at its lowest point (θ = 0).

Question

 What is the gravitational potential energy, in joules, of the ball before it is released? 

What will be the speed of the ball, in meters per second, when it reaches the bottom? 

Answer 1

It really helps if you take a straight object like a pencil or some other household object and use that to demonstrate what is going on in the problem. When you swing the pencil like a pendulum back to a point that is roughly 28 degrees from vertical, you can see that the change in height of the ball is small. Mathematically, the height change is:

h = L - Lcos(θ)

where h = height , L = length of the rope, and θ = angle with respect to vertical. Next, gravitational potential energy is given by the following equation:

U = mgh

where m = mass of the object, g = acceleration due to gravity, and h = height which we now have an expression for. The gravitational potential energy is now:

U = mg(L - Lcos(θ))

U = (0.25 kg)(9.81 m/s²)(0.85m - 0.85cos(28^o)m)

U = 0.244 J

To determine the velocity of the object at the bottom of its motion, all of the energy has gone from gravitational potential into kinetic since at the bottom, the problem says that U = 0. The kinetic energy of an object is given by the following equation:

KE = (1/2)mv²

where m = mass of the object and v = velocity of the object. Since we know that all of the energy was transferred into kinetic energy at the bottom, we can conclude that:

0.244 J = (1/2)mv²

After this step, you solve the equation for the velocity which gives the following expression:

v² = (2*0.244 J)/m

So when you plug the numbers into the expression, you should get:

v² = (2*0.244 J)/(0.25 kg)

v² = 1.952 m²/s²

v = 1.40 m/s

and that is the final answer for the speed of the ball at the bottom!