Question

A particle starting from rest moves in a circle of radius r,it attains a velocity Vo in the nth round. find its angular acceleration
a) Vo/n rad/s²
b) Vo²/2πnr² rad/s²
c) Vo²/4πnr² rad/s²
d) Vo²/4πnr rad/s²

Answer 1

Answer:

C

Explanation:

$\displaystyle \theta = 2\pi n$

$\displaystyle \omega = \frac{v_0}{r}$

$\omega_f^2 - \omega_i^2 = 2\alpha \theta$

$\displaystyle \left(\frac{v_0}{r}\right)^2 = 2\alpha\times 2\pi n$

$\displaystyle \therefore \alpha = \frac{v_0^2}{4\pi nr^2}$

Answer 2

The angular acceleration of the particle is, $\alpha =\dfrac{v_o^2}{4\pi nr^2}\ rad/s^2$

Explanation:

A particle starting from rest moves in a circle of radius r,it attains a velocity Vo in the nth round. The angular acceleration is given by the change in angular velocity per unit time.

For a rotation, $\theta=2\pi n$

Also, $\omega_f=\dfrac{v}{r}$

Using equation of motion as :

$\omega_f^2=2\alpha \theta$

$\alpha =\dfrac{\omega_f^2}{2\theta}$

$\alpha =\dfrac{(v_o/r)^2}{2\times 2\pi n}$

$\alpha =\dfrac{v_o^2}{4\pi nr^2}\ rad/s^2$

So, the correct option is (c).

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