Question

A particle $A$ moves along a circle of radius $R=50cm$ so that its radius vector $r$ relative to the point $O$ (figure) rotates with the constant angular velocity $\omega =0.40rad/s$. Then magnitude of the velocity of the particle, and the magnitude of its total acceleration will be

• A. $v=0.4m/s,a=0.4m/s^2$
• B. $v=0.32m/s,a=0.32m/s^2$
• C. $v=0.32m/s,a=0.4m/s^2$
• D. $v=0.4m/s,a=0.32m/s^2$

Answer 1

The correct option is D $v=0.4m/s,a=0.32m/s^2$

Given,

$R=0.50m$

$\omega =0.4rad/s$

$\theta =\frac{{\theta }^{\prime }}{2}$

${\omega }^{\prime }=2\omega$

$v^{\prime }=2\omega R=2\times 0.40\times 0.50=\underset{–}{0.4m/s}$

$a={\omega }^{2}R$

$a^{\prime }={\omega }^{\prime 2}R=(2\omega {)}^2\times 0.50=\underset{–}{0.32m/s^2}$