Question

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?
A) increasing both the radius and the speed by a factor of 9
B) Keeping the speed fixed and decreasing the radius by a factor of 9
C) Keeping the radius fixed and increasing the speed by a factor of 9
D) Keeping the speed fixed and increasing the radius by a factor of 9
E) decreasing both the radius and the speed by a factor of 9
F) Keeping the radius fixed and increasing the speed by a factor of 3

Answer 1

Answer:

A, B and F are the right answers

Explanation:

The formula is $a_{c} = v^{2} / r$

A) Increasing both the radius and the speed by a factor of 9

$a_{c} = (9v)^{2} / (9r)$ = $9v^{2} / r$

which means the acceleration will increase by a factor of 9

B) Keeping the speed fixed and decreasing the radius by a factor of 9

$a_{c} = (v)^{2} / (r/9)$ = $9v^{2} / r$

which means the acceleration will increase by a factor of 9

C) Keeping the radius fixed and increasing the speed by a factor of 9

$a_{c} = (9v)^{2} / (r)$ = $81v^{2} / r$

It will increase the acceleration 81 times

D) Keeping the speed fixed and increasing the radius by a factor of 9

$a_{c} = (v)^{2} / (9r)$ = $v^{2} / 9r$

It will decrease the acceleration by 9 times

E) decreasing both the radius and the speed by a factor of 9

$a_{c} = (v/9)^{2} / (r/9)$ = $v^{2} / 9r$

It will decrease the acceleration by 9 times

F) Keeping the radius fixed and increasing the speed by a factor of 3

$a_{c} = (3v)^{2} / (r)$ = $9v^{2} / r$

It will increase the acceleration by 9 times