A bowling ball weighing 71.2 {\ m N} is attached to the ceiling by a 4.00 {\ m m

Question

A bowling ball weighing 71.2 { m N} is attached to the ceiling by a 4.00 { m m} 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 3.90 { m m}/{ m s}.

a.) At that instant, find the magnitude of the acceleration of the bowling ball (m/s2).

b.) At that instant, find the direction of the acceleration of the bowling ball. Upward, Downward, or in the direction of the motion?

c.) At that instant, find the tension in the rope(N).

Explanation / Answer

Use Newton’s 2nd law to write an equation for the net centripetal forces on the ball is it swings through its arced path:

?F = mv²/r = T - mg
T = m[(v²/r) + g]

The mass of the ball is:

m = w/g
= 71.2N / 9.80m/s²
= 7.27kg

So:

T = 7.27kg[(16.0m²/s² / 3.80m) + 9.80m/s²]
= 102N

Hope this helps.

the acceleration is centripetal and is:

a = v²/r
= 16m²/s² / 3.80m
= 4.21m/s² upward

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