You decide to make a simple pendulum by attaching a bowling ball to a very light wire and suspending it from the ceiling in your garage. The length of the wire is 1.75 m and the mass of the bowling ball is 5 kg. If you pull the bowling ball back an angle of 15 degrees from the vertical and release it, how fast will the bowling ball be moving when it passes through equilibrium?
PE of ball when released is mgh
KE of ball at bottom of swing is the same, and is 1/2 mv^2.
get PE, solve for v
To find the speed of the bowling ball when it passes through equilibrium, we can use the principle of conservation of energy.
First, let's calculate the potential energy of the system when the ball is pulled back up to an angle of 15 degrees from the vertical. The potential energy of a pendulum can be calculated using the formula:
Potential Energy = mass * gravity * height
In this case, the height is the vertical distance between the equilibrium position and the highest point the ball reaches during its swing. Since the pendulum is being pulled back by an angle of 15 degrees, the height can be calculated as:
height = length * (1 - cos(angle))
where length is the length of the wire and angle is the angle measured in radians. Let's calculate it:
angle = 15 degrees * (π/180) (converting degrees to radians)
= 0.2618 radians
height = 1.75 m * (1 - cos(0.2618))
≈ 0.0588 m
Next, we can calculate the potential energy:
Potential Energy = 5 kg * 9.8 m/s^2 * 0.0588 m
≈ 2.7112 Joules
According to the principle of conservation of energy, this potential energy will be converted to kinetic energy when the ball passes through equilibrium. The kinetic energy can be calculated as:
Kinetic Energy = 1/2 * mass * velocity^2
where mass is the mass of the ball and velocity is the speed of the ball when it passes through equilibrium.
Setting the potential energy equal to the kinetic energy, we have:
Potential Energy = Kinetic Energy
=> 2.7112 Joules = 1/2 * 5 kg * velocity^2
Simplifying, we find:
velocity^2 = (2 * 2.7112 Joules) / 5 kg
≈ 2.169 Joules / kg
Finally, taking the square root, we can find the velocity:
velocity = √(2.169 Joules / kg)
≈ 1.472 m/s
Therefore, the bowling ball will be moving at a speed of approximately 1.472 m/s when it passes through equilibrium.