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:
$\theta = 2t^3 + 0.5$
Where $\theta$ is in radians and $t$ is in seconds,then the angular velocity of the particle at $t = 2 \, s$ is:
- A$8 \, rad/s$
- B$12 \, rad/s$
- C$24 \, rad/s$
- D$36 \, rad/s$
Explore More
Similar Questions
$A$ wheel initially at rest is subjected to a uniform angular acceleration about its axis. In the first $2 \text{ s}$ it rotates through an angle $\theta_1$ and in the next $2 \text{ s}$ it rotates through an angle $\theta_2$. The ratio $\frac{\theta_2}{\theta_1}$ is . . . . . . .
$A$ particle at rest starts moving with a constant angular acceleration of $4 \ rad/s^2$ in a circular path. At what time will the magnitude of its centripetal acceleration and tangential acceleration be equal?
$A$ particle performing uniform circular motion of radius $\frac{\pi}{2} \ m$ makes $x$ revolutions in time $t$. Its tangential velocity is
When a ceiling fan is switched on,it makes $10$ rotations in the first $3\,s$. How many rotations will it make in the next $3\,s$? (Assume uniform angular acceleration)
Difficult
View Solution$A$ disc starts from rest and acquires an angular velocity of $240 \ rpm$ in $10 \ s$. The angular acceleration is ...... $rad/s^2$ (assuming it is constant).
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo