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,

$a=\frac{v^2}{r}\ldots\ldots(1)$

The linear velocity of the body in circular motion is,

$v=\frac{2\pi r}{T}\ldots\ldots(2)$

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

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

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

$T^2=\frac{4\pi^2r}{a}$

Substituting the known values,

$\begin{gathered} T^2=\frac{4\pi^2\times1.5}{2.5} \\ T^2=23.69 \\ T\approx4.9\text{ s} \end{gathered}$

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