A bowling ball weighing 71.2 is attached to the ceiling by a 4.00 rope. The ball

Question

A bowling ball weighing 71.2 is attached to the ceiling by a 4.00 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 .


At that instant, find the magnitude of the acceleration of the bowling ball.
=




Part B
At that instant, find the direction of the acceleration of the bowling ball.
Upward
Downward
In the direction of motion


Part C
At that instant, find the tension in the rope.
=

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. Later noticed you needed the acceleration as well. It is centripetal and is: a = v²/r = 16m²/s² / 3.80m = 4.21m/s²

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.