Question

In uniform circular motion, how does the acceleration change when the speed is increased by a factor of 3? When the radius is decreased by a factor of 2?

Answer 1

Answer:

The acceleration will become 9/2 times.

a' =9/2 a

Explanation:

We know that acceleration of a particle when it is moving in the circular path is given as

$a=\omega^2\ r$

r=radius

ω= angular speed

If the speed ω '= 3 ω

If the radius ,$r'=\dfrac{r}{2}$

The final acceleration =a'

$a'=\omega^2'\ r'$

$a'=(3\omega)^2\times \dfrac{r}{2}$

$a'=9\omega^2\times \dfrac{r}{2}$

$a'=\omega^2\times \dfrac{9r}{2}$

$a'= \dfrac{9r}{2}\times \omega^2\times r$

$a'=\dfrac{9}{2}a$

Therefore the acceleration will become 9/2 times.