Respuesta :
Explanation:
When an object moves in a circular path, it will have circular acceleration. Its magnitude of acceleration is given by :
[tex]a=\omega^2R[/tex]
Since, [tex]\omega=\dfrac{2\pi }{T}R[/tex]
[tex]a=(\dfrac{2\pi}{T})^2R[/tex]
T is the time period
R is the radius of the circular path
To increase the centripetal acceleration bu a factor of 1.5 or 3/2, radius of circle must be increase by a factor of 6 and T is increased by a factor of 2 such that,
R'=6R and T'=2T
So,
[tex]a'=(\dfrac{2\pi}{T'})^2R'[/tex]
[tex]a'=(\dfrac{2\pi}{(2T)})^2(6R)[/tex]
[tex]a'=\dfrac{6}{4}(\dfrac{2\pi}{T})^2R[/tex]
[tex]a'=\dfrac{3}{2}(\dfrac{2\pi}{T})^2R[/tex]
Hence, this is the required solution.
