Question

A car of mass $m$ is moving on a horizontal circular track of radius $r$. The time taken to finish one complete revolution is $\frac{\pi }{2}\text{s}$. If the velocity of the car at an instant $t$ is given by $v=5\sin 2t$, then the magnitude of tangential acceleration of the car at the time when it finishes a quarter circle is $p$. The value of $p^2$ is (No slipping between the car and track)

• A. 6.3
• B. 6.30
• C. 6.2
• D. 6.25
• E. 6.20

Answer 1

Answer: (A), (B), (C), (D), (E)

Given, $v=5\sin 2t$
$\Rightarrow \frac{dx}{dt}=5\sin 2t$ where $x$ is the distance covered
$\Rightarrow {\int }_0^{x}dx={\int }_0^{t}5\sin 2tdt$
$\Rightarrow x=-\frac{5}{2}\cos 2t..\left(1\right)$
Given, time period for one complete revolution $=\frac{\pi }{2}s$
i.e for $t=\frac{\pi }{2}s,x=2\pi r$
$\Rightarrow 2\pi r=\frac{5}{2}\Rightarrow r=\frac{5}{4\pi }$
So, for quarter revolution, $x=\frac{\pi r}{2}=\frac{5}{8}$ Therefore, $\cos 2t=-\frac{1}{4}$

Tangential acceleration $a_t=\frac{dv}{dt}=\frac{d}{dt}\left(5\sin 2t\right)$
$\Rightarrow a_t=5\times 2\cos 2t$
Substituting $\cos 2t=-\frac{1}{4}$
$\Rightarrow a_t=5\times \frac{-1}{2}=-2.5$
$\Rightarrow p=2.5$
$\therefore$ The value of $p^2={2.5}^2=6.25\text{units}$