Question

A particle moves along a circle of radius $\frac{20}{\pi }m$ with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begin the tangential acceleration is

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

Answer 1

The correct option is D $40m/s^2$

$\begin{array}{c}Accordingtoqestion..................\\ Initialangularvelocity:{\omega }_0=0\\ Finalangularvelocity:\omega =\frac{80}{20/\pi }=4\pi rad/s\left[\omega =\frac{v}{r}\right]\\ Angulardisplacement:\theta =2\left(2\pi \right)=\left(4\pi \right)rad.\\ by{\omega }^2={\omega }_0^2+2a\theta \\ Now,\\ \Rightarrow {\left(4\pi \right)}^2=0^2+2a\left(4\pi \right)\left[\begin{array}{c}thelinearvariablesbasedequ=\\ v^2=u^2+2as\end{array}\right]\\ \Rightarrow 16{\pi }^2=8a\pi \\ \therefore a=2\pi rad/s^2\\ Also,relationbetweenangularandtangentialacceleration:\left(a_t=ar\right)\\ \Rightarrow a_t=\left(2\pi \right)\left(\frac{20}{\pi }\right)\\ \therefore a_t=40m/s^2\\ \\ sothatthecorrectoptionisD.\end{array}$