Question

A bowling ball weighing 71.7 N is attached to the ceiling by a rope of length 3.73 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 4.60 m/s. At this instant, what are:
a. the acceleration of the bowling ball, in magnitude and direction
b. the tension in the rope?

Answer 1

Answer:

A) a = 5.673 m/s²

The direction will be upwards vertically towards the point where it is suspended.

B) T = 113.2 N

Explanation:

A) We are given;

Weight of bowling ball; W = 71.7 N

Speed; v = 4.6 m/s

Rope length; r = 3.73 m

Now, formula for the centripetal acceleration is;

a = v²/r

Thus; a = 4.6²/3.73

a = 5.673 m/s²

The direction will be upwards vertically towards the point where it is suspended.

B) since weight is 71.7 N, it means that;

Mass = weight/acceleration = 71.7/9.8

Mass(m) = 7.316 kg

Thus,

Centripetal force is;

F_cent = 7.316 × 5.673

F_cent = 41.5 N

Thus, Tension in the rope is;

T = W + F_cent

T = 71.7 + 41.5

T = 113.2 N