## A steel ball is whirled on the end of a chain in a horizontal circle of radius R with a constant period T. If the radius of the circle is th

Question

A steel ball is whirled on the end of a chain in a horizontal circle of radius R with a constant period T. If the radius of the circle is then reduced to 0.75R, while the period remains T, what happens to the centripetal acceleration of the ball?

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1 year 2021-09-04T15:03:40+00:00 1 Answers 11 views 0

The centripetal acceleration will be increased to 1.33 of its initial state.

Explanation:

### Centripetal acceleration

The Centripetal acceleration of  an object is the acceleration of the object along a circular path  moving towards the center of the circular path. The centripetal acceleration is represented in the equation bellow

$$a_{c} = \frac{V^{2} }{r}$$ ……………………………….. 1

where $$a_{c}$$ is the centripetal acceleration

v is the tangential velocity

Reference to equation 1 the centripetal acceleration ( $$a_{c}$$) is inversely proportional ($$y = \frac{k}{x}$$) to the radius of the circle or path. this means that when the radius increases the centripetal acceleration reduces and when the radius reduces the centripetal acceleration increases. The radius was reduced to 0.75R in the question that will amount to 1.33$$a_{c}$$ increase in the centripetal acceleration. This can be obtained by multiplying the centripetal acceleration by the inverse of 0.75 which is 1.33.
Therefore, when the radius is reduced by 0.75R , the centripetal acceleration of the steel ball will increase by 1.33$$a_{c}$$. since the period is kept constant