If an object is performing a uniform circular motion with a constant angular speed of $20 r a d / s$ . Find the angular displacement in $2 s e c$ .

20 rad

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40 rad

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

60 rad

No worries! We‘ve got your back. Try BYJU‘S free classes today!

80 rad

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
40 rad

Given,

Angular speed $ω$ = $20 r a d / s$
Time $t$ = $2 s e c$
Angular displacement, $θ = ω t$
$= 20 \times 2$ = $40 r a d$

Suggest Corrections

Similar questions

Q. Angular velocity of a particle performing circular motion is defined as the time rate of change of limiting angular displacement.

Q. A particle performs uniform circular motion with an angular momentum L. If the frequency of particle's motion is doubled and its kinetic energy halved, the angular momentum becomes:

A particle performs uniform circular motion with an angular momentum L. If the angular frequency f of the particle is doubled, and kinetic energy is halved, its angular momentum becomes

Q. A particle initially at rest performs circular motion with uniform angular acceleration 0.2 rad/

s^{2}

. What speed will it attain in 10 seconds?

Q. The angular displacement of an object having uniform circular motion is

\frac{π}{4}

rad in every 3 s.Find the frequency of revolution.

Join BYJU'S Learning Program

If an object is performing a uniform circular motion with a constant angular speed of 20 rad/s. Find the angular displacement in 2 sec.

The correct option is B 40 rad Given, Angular speed ω = 20 rad/s Time t = 2 sec Angular displacement, θ=ωt =20×2= 40 rad

If an object is performing a uniform circular motion with a constant angular speed of $20 r a d / s$ . Find the angular displacement in $2 s e c$ .

The correct option is B
40 rad

Given,

Angular speed $ω$ = $20 r a d / s$
Time $t$ = $2 s e c$
Angular displacement, $θ = ω t$
$= 20 \times 2$ = $40 r a d$