A bowling ball weighing 71.3 is attached to the ceiling by a rope of length 3.73 . 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 4.60 .
What is the acceleration of the bowling ball, in magnitude and direction, at this instant?
To find the acceleration of the bowling ball at the instant when it passes through the vertical, we can use the equation for centripetal acceleration:
a = (v^2) / r
Where:
a = acceleration
v = velocity
r = radius of the circular path
In this case, the radius of the circular path is the length of the rope, which is 3.73 m. We are given that the speed of the bowling ball is 4.60 m/s.
Substituting the given values into the equation:
a = (4.60^2) / 3.73
a ≈ 5.673 m/s^2
Therefore, the magnitude of the acceleration of the bowling ball at this instant is approximately 5.673 m/s^2.
Now, let's determine the direction of the acceleration. Since the bowling ball is passing through the vertical, the acceleration is directed towards the center of the circular path, which is downwards.
Hence, the magnitude of the acceleration is approximately 5.673 m/s^2, and its direction is downwards.