A bowling ball weighing 71.2 is attached to the ceiling by a 3.90 rope. The ball is pulled to one side and released; it then swings back and forth like a pendulum. As the rope swings through its lowest point, the speed of the bowling ball is measured at 4.20 .
Please complete the question. What is required?
Also the quantities do not indicate units, so it is not clear if the quantities are in consistent units.
If they are in consistent units, the ball had been pulled aside to 13.34° with the vertical.
To calculate the speed of the bowling ball at its lowest point, we can use the principle of conservation of energy.
First, let's find the potential energy of the bowling ball at its highest point (when it is pulled to one side and released). The potential energy (PE) is given by the equation:
PE = m * g * h
where m is the mass of the bowling ball (in kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the lowest point (in meters).
Since the bowling ball is initially at rest at its highest point, all of its potential energy is converted into kinetic energy at its lowest point. Therefore, the kinetic energy (KE) of the bowling ball at its lowest point is equal to its potential energy at its highest point:
KE = PE
Next, let's find the kinetic energy of the bowling ball at its lowest point. The kinetic energy (KE) is given by the equation:
KE = (1/2) * m * v^2
where v is the speed of the bowling ball (in m/s).
Now, we can set the potential energy equal to the kinetic energy and solve for the speed (v):
m * g * h = (1/2) * m * v^2
Since the mass (m) cancels out on both sides of the equation, we can simplify it to:
g * h = (1/2) * v^2
Solving for v, we get:
v = √(2 * g * h)
Now, substituting the given values into the equation, we have:
v = √(2 * 9.8 * 3.90)
v ≈ 8.81 m/s
Therefore, the speed of the bowling ball at its lowest point is approximately 8.81 m/s.