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 $\omega =0.4rad/s$. Then modulus of the velocity of the particle, and the modulus of its total acceleration will be:

• A. $v=0.02m/s,a=0.008m/s^2$
• B. $v=0.32m/s,a=0.32m/s^2$
• C. $v=0.32m/s,a=0.04m/s^2$
• D. $v=0.4m/s,a=0.32m/s^2$

Answer 1

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

We fix the coordinate system as shown so that $\angle BOA=\theta$

The particles rotates clockwise thus $w=-d\theta /dt$

From triangle $OAB$

$\frac{R}{\sin \theta }=\frac{r}{\sin (\pi -2\theta )}\Rightarrow r=2R\cos \theta$

Since

$\overrightarrow{r}=r\cos \theta \hat{i}+r\sin \theta \hat{j}=2R{cos}^2\theta \hat{i}=Rsin2\theta \hat{j}\frac{d\overrightarrow{r}}{dt}=\overrightarrow{v}=2R2cos(-sin\theta )\frac{d\theta }{dt}\hat{i}+2Rcos2\theta \frac{d\theta }{dt}\hat{j}=2Rw(sin2\theta \hat{i}-{cos}^2\theta \hat{j})\left|\overrightarrow{v}\right|=2wR=0.4m/s$

Since $w$ is constant $\frac{dv}{dt}=0$

$\therefore$ centripetal acceleration:

$a_c=\frac{v^2}{R}=4w^{2}R=0.32m/s^2$