A particle moves along a circle of radius lef | vedclass

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

Question

(A) Given: Radius $r = \frac{20}{\pi} \, m$,initial angular velocity $\omega_i = 0$,and total angular displacement $	heta = 2 	imes (2 \pi) = 4 \pi

Vedclass · Accepted Answer

$40 \, m/s^2$

Vedclass · Answer

(A) Given: Radius $r = \frac{20}{\pi} \, m$,initial angular velocity $\omega_i = 0$,and total angular displacement $	heta = 2 	imes (2 \pi) = 4 \pi \, rad$ (since it completes two revolutions).Final linear velocity $v = 80 \, m/s$.The final angular velocity is $\omega_f = \frac{v}{r} = \frac{80}{20/\pi} = 4 \pi \, rad/s$.Using the rotational kinematic equation $\omega_f^2 = \omega_i^2 + 2 \alpha 	heta$,we get:$(4 \pi)^2 = 0^2 + 2 \alpha (4 \pi)$$16 \pi^2 = 8 \pi \alpha$$\alpha = 2 \pi \, rad/s^2$.The tangential acceleration is $a_t = r \alpha = \left( \frac{20}{\pi} ight) 	imes (2 \pi) = 40 \, m/s^2$.

$A$ 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 begun,the tangential acceleration is

Similar Questions

The angular speed of a flywheel moving with uniform angular acceleration changes from $1200\,rpm$ to $3120\,rpm$ in $16\,s$. The angular acceleration in $rad/s^{2}$ is: (in $\pi$)

$A$ particle performing uniform circular motion of radius $\frac{\pi}{2} \ m$ makes $x$ revolutions in time $t$. Its tangential velocity is

If the equation for the angular displacement of a particle moving on a circular path is given by:
$\theta = 2t^3 + 0.5$
Where $\theta$ is in radians and $t$ is in seconds,then the angular velocity of the particle at $t = 2 \, s$ is:

$A$ wheel is at rest. Its angular velocity increases uniformly and becomes $60 \ rad/sec$ after $5 \ sec$. The total angular displacement is ........ $rad$.

$A$ ball is spun with angular acceleration $\alpha = 6t^2 - 2t$,where $t$ is in seconds and $\alpha$ is in $rad/s^2$. At $t = 0$,the ball has an angular velocity of $10 \ rad/s$ and an angular position of $4 \ rad$. The most appropriate expression for the angular position of the ball is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module