$A$ cycle wheel of radius $0.4\ m$ completes one revolution in one second. Then,the acceleration of a point on the rim of the cycle wheel will be:

  • A
    $0.8\ m/s^2$
  • B
    $0.4\ m/s^2$
  • C
    $1.6\pi^2\ m/s^2$
  • D
    $0.4\pi^2\ m/s^2$

Explore More

Similar Questions

For a particle in uniform circular motion:

Write the relation between linear velocity and angular velocity.

Difficult
View Solution

For uniform circular motion,write the equations for centripetal force and centripetal acceleration. Also,provide examples.

The position vector of a particle is $\vec{r} = a \cos \omega t \hat{i} + a \sin \omega t \hat{j}$. Calculate the angle between its position vector and the velocity vector. $(\cos 0 = 1, \cos 90 = 0, \cos 60 = 0.5, \sin 30 = 0.5, \sin 0 = 0)$ (in $^\circ$)

Assertion $(A)$: The speed of a body in uniform circular motion is constant.
Reason $(R)$: In uniform circular motion,the acceleration of the body is constant.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo