Question

A man is whirling a 0.25 kg ball on a 1.5 m long string at 3 m/s. Find the centripetal acceleration of this ball.
O 13.5 m/s2
0 0.5 m/s2
m.
2 m/s2
6 m/s2

Answer 1

The centripetal acceleration of the ball been whirled in a circular path is 6m/s².

Hence, Option D) 6m/s² is the

Given the data in the question:

• Mass of ball; $m = 0.25kg$

• Length of string or radius; $r = 1.5m$

• Speed; $v = 3m/s$

Centripetal acceleration is the acceleration of a particle moving in a circular path. It is expressed as:

$a_c = \frac{v^2}{r}$

Where $a_c$ is the Centripetal acceleration, v is the speed of the motion and r is the length of the string or radius of the circular pattern.

We substitute our given values into the equation

$a_c = \frac{(3m/s)^2}{1.5m}\\\\a_c = \frac{9m^2/s^2}{1.5m}\\\\a_c = 6m/s^2$

The centripetal acceleration of the ball been whirled in a circular path is 6m/s².

Hence, Option D) 6m/s² is the correct answer.

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