Question

A spinning top rotates so that its edge turns at a constant speed of 0.63 m/s. If the top has a radius of 0.10 m, what is the

Answer 1

Answer:

The answer is 4.0 m/s2

Answer 2

Answer:

The  centripetal acceleration of the edge of the top $6.3m/s^2$

Explanation:

Given:

Speed with which the top rotates =  0.63 m/s

Radius of the top = 0.10 m

To Find:

Centripetal acceleration of the edge of the top =?

Solution:

The centripetal acceleration is the rate at which the tangential velocity changes. Also its direction is always inwards along the radius vector of the circular motion.

We know that centripetal acceleration is

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

Where

v = velocity

r = radius

Substituting the values

$a_c = \frac{0.63}{0.10}$

$a_c = 6.3$$m/s^2$