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
[tex]a=\omega^2\ r[/tex]
r=radius
ω= angular speed
If the speed ω '= 3 ω
If the radius ,[tex]r'=\dfrac{r}{2}[/tex]
The final acceleration =a'
[tex]a'=\omega^2'\ r'[/tex]
[tex]a'=(3\omega)^2\times \dfrac{r}{2}[/tex]
[tex]a'=9\omega^2\times \dfrac{r}{2}[/tex]
[tex]a'=\omega^2\times \dfrac{9r}{2}[/tex]
[tex]a'= \dfrac{9r}{2}\times \omega^2\times r[/tex]
[tex]a'=\dfrac{9}{2}a[/tex]
Therefore the acceleration will become 9/2 times.
