Question

Need help with promblem d, I think I have the correct formula for this example.

Answer 1

Given data:

* The centripetal acceleration of the system is,

`a=2.5ms^(-2)`

* The radius of the circular motion is r = 1.5 m.

Solution:

The centripetal acceleration in terms of the linear velocity is,

`a=(v^2)/(r)\ldots\ldots(1)`

The linear velocity of the body in circular motion is,

`v=(2\pi r)/(T)\ldots\ldots(2)`

Substituting the value of velocity from the (2) equation to the (1) in the equation,

`\begin{gathered} a=(((2\pi r)/(T))^2)/(r) \\ a=(4\pi^2r^2)/(T^2r) \\ a=(4\pi^2r)/(T^2) \end{gathered}`

Thus, the time period in terms of the centripetal acceleration and radius is,

`T^2=(4\pi^2r)/(a)`

Substituting the known values,

`\begin{gathered} T^2=(4\pi^2*1.5)/(2.5) \\ T^2=23.69 \\ T\approx4.9\text{ s} \end{gathered}`

Hence, the time period to complete one revolution is 4.9 seconds.