First, we can find the mass of the bowling ball using its weight:
m = 71.8 N / 9.81 m/s^2
m = 7.32 kg
Next, let's find the centripetal acceleration of the bowling ball as it swings through the vertical.
ac = v^2 / r
ac = (4.50 m/s)^2 / 3.83 m
ac = 5.31 m/s^2
The centripetal acceleration is directed towards the center of the circle, so at the instant when the rope is vertical, the centripetal acceleration is upwards.
Next, we can find the tension in the rope at this instant. The tension must provide two forces: one to balance the weight of the bowling ball and the other to provide the centripetal acceleration. So, the tension can be calculated as:
T = m * (g + ac)
T = 7.32 kg * (9.81 m/s^2 + 5.31 m/s^2)
T = 7.32 kg * 15.12 m/s^2
T = 110.68 N
So at that instant, the acceleration of the bowling ball is 5.31 m/s^2 upwards, and the tension in the rope is 110.68 N.
A bowling ball weighing 71.8 N is attached to the ceiling by a rope of length 3.83 m. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.50 m/s.
What is the acceleration of the bowling ball, in magnitude and direction, at this instant?
What is the tension in the rope at this instant?
1 answer