Question

A small body of mass $m=0.10kg$ moves in the reference frame rotating about a stationary axis with constant angular velocity $\omega =5.0rad/s$. What work does the centrifugal force of inertia perform during the transfer of this body along an arbitrary path from point $1$ to point $2$ which are located at the distance $r_1=30cm$ and $r_2=50cm$ from the axis of rotation?

• A. $0.15J$
• B. $0.4J$
• C. $0.2J$
• D. $0.35J$

Answer 1

The correct option is C $0.2J$

Given mass, $m=0.1kg$
Angular velocity, $\omega =5rad/s$
Considering the rotating frame of reference X-Y rotating about the origin O.

Let the body covers arbitrary path $AB$ as shown above while moving in he rotating frame of reference where $r_1$ and $r_2$ represents the position vector of starting and end point.
Considering a small element $ds$ of the path $AB$. Let the element $ds$ makes an angle $\theta$ with the centrifugal force vector directed along the radial direction as shown below.
Work done by centrifugal force to move the body by displaccement $ds$ will be, $dW=\overrightarrow{F}.\overrightarrow{ds}$
$\begin{array}{}dW=Fdscos\theta & \text{(1)}\end{array}$
Centrifugal force, $F=m{\omega }^{2}r$
$\begin{array}{}\Rightarrow dW=m{\omega }^{2}rdscos\theta & \text{(2)}\end{array}$
Now radial vector component for the displacement $ds$ will be,
$dr=dscos\theta$
Therefore equation $\left(2\right)$ becomes,
$dW=m{\omega }^{2}rdr$
Integrating the above equation from $r_1$ to $r_2$,
$\int dW={\int }_{r_1}^{r_2}m{\omega }^{2}rdr$
$W=m{\omega }^2[\frac{r^2}{2}{]}_{0.3}^{0.5}$
$\Rightarrow W=0.1\times 5^2\times \left[\frac{0.25-0.09}{2}\right]$
$\Rightarrow W=0.2$ J

For detailed solution watch next video.