Looking for classes? Ksquare Career Institute, Bengaluru →
A particle moves along a circle of radius $\left(\dfrac{20}{\pi}\right)$ 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 begun, the tangential acceleration is : -
A
40 m s$^{-2}$
B
$640\pi$ m s$^{-2}$
C
$160\pi$ m s$^{-2}$
D
$40\pi$ m s$^{-2}$
Detailed Solution
The tangential acceleration is constant, $\dfrac{dv}{dt} = a$ (say), so the equations of uniformly accelerated motion apply along the circular path.
Use $v^2 = u^2 + 2as$, where s is the distance travelled along the circle.
Distance in two revolutions: $s = 2 \times 2\pi r = 2 \times 2\pi \times \dfrac{20}{\pi} = 80$ m
$u = 0$ (motion begins from rest), $v = 80$ m/s
$80^2 = 0 + 2 \times a \times 80$
$a = \dfrac{6400}{160} = 40$ m s$^{-2}$
So the tangential acceleration is 40 m s$^{-2}$.
Use $v^2 = u^2 + 2as$, where s is the distance travelled along the circle.
Distance in two revolutions: $s = 2 \times 2\pi r = 2 \times 2\pi \times \dfrac{20}{\pi} = 80$ m
$u = 0$ (motion begins from rest), $v = 80$ m/s
$80^2 = 0 + 2 \times a \times 80$
$a = \dfrac{6400}{160} = 40$ m s$^{-2}$
So the tangential acceleration is 40 m s$^{-2}$.