Question

A bowling ball is on the end of a rope of length 5 meters, the other end of which is attached to a hook in the ceiling. The ball is displaced to one side so that the rope is horizontal, then released so that it swings down through the vertical to the other side. What is the magnitude of its acceleration as it moves through its lowest point

Answer 1

Answer:

$a = 0\,\frac{m}{s^{2}}$

Explanation:

Let consider the following system, which is described in the image attached below and two reference axis, one parallel and the other perpendicular to the direction of motion. The corresponding equations of equilibrium are described herein:

$\Sigma F_{n} = T - m\cdot g \cdot \cos \theta = 0$

$\Sigma F_{t} = m\cdot g \cdot \sin \theta = m\cdot a$

The acceleration of the bowling ball at the lowest point occurs at $\theta = 0^{\textdegree}$

$m\cdot g \cdot \sin \theta = m\cdot a$

$g\cdot \sin \theta = a$

$a = 0\,\frac{m}{s^{2}}$