Question

The total acceleration of a particle in circular motion is $25m{s}^{-2}.$ If the speed of the particle decreases at a rate of $15m{s}^{-2}.$ If radius of the path of the particle is $r=2m$, the angular speed $\omega$ of the particle relative to the centre of the circle is

• A. $1\frac{rad}{s}$
• B. $\sqrt{10}\frac{rad}{s}$
• C. $2\frac{rad}{s}$
• D. $2\sqrt{10}\frac{rad}{s}$

Answer 1

The correct option is B $\sqrt{10}\frac{rad}{s}$

Let the tangential acceleration be $a_t$,
centripetal acceleration be $a_c$,
Net acceleration be $a$
Given,
$a=25m{s}^{-2},a_t=15m{s}^{-2},r=2m$
$a=\sqrt{a_t^2+a_c^2}\Rightarrow a^2=a_1^2+a_c^2$
$\Rightarrow {25}^2={15}^2+a_c^2$
$\therefore a_c^2={20}^2\therefore a_c=20$

$\Rightarrow {\omega }^{2}r=20$
$\Rightarrow {\omega }^2=\frac{20}{2}=10\therefore \omega =\sqrt{10}\frac{rad}{s}$