Question

g A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Which way(s) can increase the centripetal acceleration of the ball by a factor of 9? Group of answer choices Keeping the speed fixed and decreasing the radius by a factor of 9 Keeping the radius fixed and increasing the speed by a factor of 3 decreasing both the radius and the speed by a factor of 9 Keeping the radius fixed and increasing the speed by a factor of 9 increasing both the radius and the speed by a factor of 9 Keeping the speed fixed and increasing the radius by a factor of 9

Answer 1

Explanation:

The centripetal acceleration of the ball that is whirled on the end of a string in a horizontal circle of radius R at constant speed v, is given by :

$a=\dfrac{v^2}{R}$

Option (1) : Keeping the speed fixed and decreasing the radius by a factor of 9

$a=\dfrac{v^2}{R/9}$

$a=\dfrac{9v^2}{R}$

The centripetal acceleration of the ball by a factor of 9.

Option (2) : Keeping the radius fixed and increasing the speed by a factor of 3

$a=\dfrac{(3v)^2}{R}$

$a=\dfrac{9v^2}{R}$

Acceleration increases.

Option (3) : Decreasing both the radius and the speed by a factor of 9.

$a=\dfrac{(v/9)^2}{R/9}$

$a=\dfrac{(v)^2}{9R}$

Acceleration decreases by a factor of 9.

Option (4) : Keeping the radius fixed and increasing the speed by a factor of 9

$a=\dfrac{(3v)^2}{R}$

$a=\dfrac{9v^2}{R}$

Acceleration increases.

Option (5) : Increasing both the radius and the speed by a factor of 9

$a=\dfrac{(9v)^2}{9R}$

$a=\dfrac{9v^2}{R}$

Acceleration increases.

Option (6) : Keeping the speed fixed and increasing the radius by a factor of 9

$a=\dfrac{(v)^2}{9R}$

$a=\dfrac{9v^2}{R}$

Acceleration increases.

Hence, this is the required solution.