A particle moves along a circle of radius&nbs | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$a_{t}=r \alpha=$ const. $\Rightarrow \alpha=$ const.

${\omega _i} = 0$          $	heta=2 \pi$

$\omega_{f}=\frac{V}{r}

Vedclass · Accepted Answer

$40$

Vedclass · Answer

$a_{t}=r \alpha=$ const. $\Rightarrow \alpha=$ const.

${\omega _i} = 0$          $	heta=2 \pi$

$\omega_{f}=\frac{V}{r}=\frac{80}{(20 / \pi)}=4 \pi$

$\alpha=\frac{\omega_{f}^{2}-\omega_{i}^{2}}{2 	heta}=2 \pi \operatorname{rad} / \sec ^{2}$

$\mathrm{a}_{\mathrm{t}}=\frac{20}{\pi} 	imes 2 \pi=40 \mathrm{\,m} / \mathrm{s}^{2}$

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

Similar Questions

For a particle in uniform circular motion, the acceleration $\vec a$ at a point $P(R,\theta)$ on the circle of radius $R$ is (Here $\theta$ is measured from the $x-$ axis)

In the given figure, $a = 15 \,m s^{- 2}$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R = 2.5\, m$ at a given instant of time. The speed of the particle is ........ $m/s$

A scooter is going round a circular road of radius $100 \,m$ at a speed of $10 \,m/s$. The angular speed of the scooter will be ......... $rad/s$

The angular acceleration of a body, moving along the circumference of a circle, is :

If the string of a conical pendulum makes an angle $\theta$ with horizontal, then square of its time period is proportional to