Question

A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diameter1.0 m.

Answer 1

Complete question:

A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diameter 1.0 m. The ball makes 2.0 revolutions every 1.0 s. What are the magnitude and direction of the acceleration of the ball?

Answer:

The acceleration of the ball is 78.98 m/s², directed inwards

Explanation:

Given;

mass of the ball, m = 175 g

radius of the circle, r = 0.5 m

angular speed of the ball, ω = 2 rev/s

The magnitude of the centripetal acceleration of the ball is calculated as follows;

$a_c = \omega^2 r\\\\where;\\\omega \ is \ angular \ speed \ in \ rad/s\\\\a_c = (2\ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} )^2 \times (0.5 \ m)\\\\a_c =78.98 \ m/s^2$

The centripetal acceleration is directed inwards.