Question

A particle moves along a circle of radius $\left(\frac{20}{\pi }\right)m$ with constant tangential acceleration. If velocity of the particle is $80\text{m/s}$ at the end of the $2^{nd}$ revolution after the motion has begun, the tangential acceleration is

• A. $40{\text{m/s}}^2$
• B. $640{\text{m/s}}^2$
• C. $160{\text{m/s}}^2$
• D. $40\pi {\text{m/s}}^2$

Answer 1

The correct option is A $40{\text{m/s}}^2$

We know,
$a_t=R\alpha {\omega }_0=0\theta =4\pi$

We can use,
${\omega }^2={\omega }_0^2+2\alpha \theta$
${\omega }^2=0+2\alpha \theta$

But, we know
$\omega =\frac{v}{R}=\frac{80}{20/\pi }=4\pi rad/s$

$\therefore (4\pi {)}^2=2\alpha 4\pi$
$\Rightarrow \alpha =2\pi$

$\Rightarrow a_t=R\alpha =\frac{20}{\pi }\times 2\pi =40m/s^2$