Angular speed of particle undergoing circular motion of radius 2 m varies with times as - -2t -1 rad-s- Distance travelled by particle in 0 to 1 s is

Answer (2) ω=2t+1 dθdt=2t+1 ⇒θ=∫_0^1(2t+1)dt ⇒θ = 2 radian Distance = θ r = 4 m ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Correlation to Linear Motion

Angular speed...

Question

Angular speed of particle undergoing circular motion of radius $2 m$ varies with time(s) as $ω = (2 t + 1)$ rad/s. Distance travelled by particle in $0$ to $1 s$ is

3 m

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

1 m

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

4 m

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

2 m

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

Open in App

Solution

The correct option is C $4 m$
Answer (2)
$ω = 2 t + 1$
$\frac{d θ}{d t} = 2 t + 1$
$\Rightarrow θ = 1 \int 0 (2 t + 1) d t$
$\Rightarrow θ = 2 radian$
Distance $= θ r$
$= 4 m$

Suggest Corrections

Similar questions

Q. Angular position of a body undergoing circular motion varies with time as

θ = 2 t^{2} + 3

radians. If radius of circular path is

1 m

, the acceleration of body at

t = 1 s

will be

Angular position of a particle performing circular motion is given by $θ = (t^{2} - 2 t) r a d$ . Consider the ads circular track $2 m$ Angular distance of particle at $t = 3 s$

Q. For a particle moving in uniform circular motion of radius

R

with angular speed

ω

, the displacement in time interval

t

2 t

Q. A particle undergoes uniform circular motion. The velocity and angular velocity of the particle at an instant of time are

\to v = 3^i + 4^j m/s

and

\to ω = x^i + 6^j rad/s

. The value of

x

rad/s

Q. The total acceleration of a particle in circular motion is

5 m / s^{2}

. If the speed of the particle decreases at a rate of

3 m / s^{2}

, assuming

r = 2 m

, the angular speed

ω

of the particle relative to the centre of the circle is

Join BYJU'S Learning Program

Angular speed of particle undergoing circular motion of radius 2 m varies with time(s) as ω=(2t+1) rad/s. Distance travelled by particle in 0 to 1 s is

The correct option is C 4 mAnswer (2) ω=2t+1 dθdt=2t+1 ⇒θ=1∫0(2t+1)dt ⇒θ=2 radian Distance =θr =4 m

Angular speed of particle undergoing circular motion of radius $2 m$ varies with time(s) as $ω = (2 t + 1)$ rad/s. Distance travelled by particle in $0$ to $1 s$ is

The correct option is C $4 m$
Answer (2)
$ω = 2 t + 1$
$\frac{d θ}{d t} = 2 t + 1$
$\Rightarrow θ = 1 \int 0 (2 t + 1) d t$
$\Rightarrow θ = 2 radian$
Distance $= θ r$
$= 4 m$