A wheel of radius 0.20m is accelerated from rest with an angular acceleration of 1 rad /s² . After a rotation of 90° the radial acceleration of a particle on its rim will be

The correct option is 0.2 π m / s².

Let us understand the step-by-step physical principles and calculations that lead to this result.

1. Identify the given parameters:
- Radius of the wheel, r = 0.20 m
- Initial angular velocity, ${\omega }_0=0\text{rad/s}$ (since the wheel starts from rest)
- Angular acceleration, $\alpha =1{\text{rad/s}}^2$
- Angular displacement, $\theta ={90}^{°}$

2. Convert the angular displacement to radians:
To perform rotational motion calculations, we convert the angle from degrees to radians:
$\theta ={90}^{°}\times \frac{\pi }{180}=\frac{\pi }{2}\text{rad}$

3. Determine the final angular velocity:
We use the rotational kinematic equation relating angular displacement, angular acceleration, and angular velocity:
${\omega }^2={\omega }_0^2+2\alpha \theta$
Substituting the known values into this equation:
${\omega }^2=0+2\times 1\times \frac{\pi }{2}$
Simplifying this gives:
${\omega }^2=\pi {\text{rad}}^2/{\text{s}}^2$

4. Calculate the radial (centripetal) acceleration:
The radial acceleration ($a_c$) of a particle on the rim of the wheel is given by:
$a_c={\omega }^{2}r$
Substituting the values of ${\omega }^2$ and $r$:
$a_c=\pi \times 0.20=0.2\pi {\text{m/s}}^2$

Thus, the radial acceleration of the particle on the rim is indeed 0.2 π m / s².