Question

How much is a ball that is tethered to a post accelerating if its linear speed is 3.5 m/s and the ball is 0.91 meters away from the pole? a = v^2 / r

Answer 1

Answer:

$a_{c} = 13.46\ m/s^2$

Explanation:

The motion of the tethered ball around the pole can be modeled as uniform circular motion. The acceleration of the objects executing uniform circular motion is due to the change in direction of their velocity. This is called centripetal acceleration. It is given by the following formula:

$a_{c} = \frac{v^2}{r}$

where,

ac = centripetal acceleration = ?

v = speed of ball = 3.5 m/s

r = distance of ball from center of rotation = 0.91 m

Using these values in equation, we get:

$a_{c} = \frac{(3.5\ m/s)^2}{0.91\ m}\\\\a_{c} = 13.46\ m/s^2$