Question

A particle $A$ moves along a circle of radius $R=50cm$ so that its radius vector $\overrightarrow{r}$ relative to the point $O$ (shown in above figure) rotates with the constant angular velocity $\omega =0.40rad/s$. Find the modulus of the velocity of the particle, and the modulus of its total acceleration.

• A. $v=2R\omega =0.40m/s$, $\alpha =4R{\omega }^2=0.32m/s^2$
• B. $v=2R\omega =0.80m/s$, $\alpha =4R{\omega }^2=0.32m/s^2$
• C. $v=2R\omega =0.40m/s$, $\alpha =4R{\omega }^2=0.64m/s^2$
• D. $v=2R\omega =0.80m/s$, $\alpha =4R{\omega }^2=0.64m/s^2$

Answer 1

Answer: (A)

$v=2R\omega =0.40m/s$, $\alpha =4R{\omega }^2=0.32m/s^2$

Let us fix the co-ordinate system at the point $O$ as shown in figure below, such that the radius vector $\overrightarrow{r}$ of point $A$ makes an angle $\theta$ with x axis at the moment shown.Note that the radius vector of the particle $A$ rotates clockwise and we here take line $ox$ as the reference line, so in this case obviously the angular velocity $\omega =(-\frac{d\theta }{dt})$ taking anti-clockwise sense of angular displacement as positive.Also form the geometry of the triangle $OAC$

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

Let us write,

$\overrightarrow{r}=r\cos {\theta }_{\overrightarrow{i}}+r\sin {\theta }_{\overrightarrow{j}}=2R{\cos }^2{\theta }_{\overrightarrow{i}}+R\sin 2{\theta }_{\overrightarrow{j}}$

Differentiating with respect to time,

$\frac{\overrightarrow{dr}}{dt}$ or $\overrightarrow{v}=2R\cos \theta (-\sin \theta )\frac{d\theta }{dt}\overrightarrow{i}+2R\cos 2\theta \frac{d\theta }{dt}\overrightarrow{j}$

or, $\overrightarrow{v}=2R\left(\frac{-d\theta }{dt}\right)[\sin 2\theta \overrightarrow{i}-\cos 2\theta \overrightarrow{j}]$

or, $\overrightarrow{v}=2R\alpha (\sin 2\theta \overrightarrow{i}-{\cos }^2\theta \overrightarrow{j})$

So, $|\overrightarrow{v}|$ or $v=2\omega R=0.4m/s$.

As $\alpha$ is constant, v is also constant and $w_t=\frac{dv}{dt}=0$,

So, $\alpha ={\alpha }_n=\frac{v^2}{R}=\frac{{\left(2\omega R\right)}^2}{R}=4{\alpha }^{2}R=0.32m/s^2$

Alternate: From the figure below, the angular velocity of the point $A$, with respect to centre of the circle $C$ becomes

$-\frac{d\left(2\theta \right)}{dt}=2\left(\frac{-d\theta }{dt}\right)=2\alpha =constant$

Thus we have the problem of finding the velocity and acceleration of a particle moving along a circle of radius R with constant angular velocity $2\alpha$.

Hence $v=2\alpha R$ and

$\alpha ={\alpha }_n=\frac{v_2}{R}=\frac{{\left(2\alpha R\right)}^2}{R}=4{\alpha }^{2}R$