Question

A particle starts moving along a circle of radius 4 m with a constant tangential acceleration of $3m/se{c}^2$. The total acceleration of the particle at t = 2 second makes an angle $\theta$ with the acceleration, where $\theta$ is equal to

• A. $ta{n}^{-}1\left(\frac{1}{3}\right)$
• B. $ta{n}^{-}1\left(3\right)$
• C. $ta{n}^{-}1\left(3\frac{1}{3}\right)$
• D. $ta{n}^{-}1\left(\frac{1}{\sqrt{10}}\right)$

Answer 1

The correct option is B $ta{n}^{-}1\left(3\right)$

Given,

That the particle starts from rest

The initial angular velocity is ${\omega }_0=0$

The radius of the circle is $r=4m$

Tangential acceleration, $a_t=3m{s}^{-2}$

Angular acceleration $\alpha =\frac{a}{r}=\frac{3}{4}rad{s}^{-2}$

$\omega ={\omega }_o+\alpha t=\frac{3}{4}\times 2=1.5rad{s}^{-1}$

$a_c=\frac{v^2}{r}={\omega }^{2}r={1.5}^2\times 4=9m{s}^{-2}$

$\theta ={\tan }^{-1}\left(\frac{a_c}{a_t}\right)={\tan }^{-1}\left(\frac{9}{3}\right)={\tan }^{-1}\left(3\right)$

Hence, angle of resultant w.r.t tangential acceleration is ${\tan }^{-1}\left(3\right)$