If the equation for the angular displacement of a particle moving on a circular path is given by theta=2t3 + 0.5 where theta is in radians and t in seconds, then the average angular velocity of the particle after 2 seconds from its start is?

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Grade 11General Physics

If the equation for the angular displacement of a particle moving on a circular path is given by theta=2t³ + 0.5 where theta is in radians and t in seconds, then the average angular velocity of the particle after 2 seconds from its start is?

Rituraj Singh

12 Years agoGrade 11

3 Answers

Arun Kumar

12 Years ago

Hello Student,

\theta=2t³ + 0.5

\theta =8.5

t=2

so average =4.25rad/s

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty

Nidhi

8 Years ago

Theta = 2t^3+0.5 ...(given)...(1)

We know that ,

Omega =d(theta)/dt

therefore,

Diff. eqn(1) wrt t,

So,

Omega= 6t^2

At t=2 ,

Omega = 6×4=24 rad/s

(Theta is displacement, Omega is angular velocity)

Hope this helps

Kushagra Madhukar

5 Years ago

Dear student,

Please find the solution to your problem below.

SInce here average angular velocity is asked and not the instantaneous angular velocity, we will have to find the total change in theta in the given 2s instead of differentiating.

Now, at t = 0, θ_o= 2(0)³ + 0.5 = 0.5 rad.

at t = 2s, θ₂= 2(2)³ + 0.5 = 8.5 rad.

Average angular velocity, w = (θ₂ – θ_o)/(2 – 0)

= 8/2

= 4 rad/s

Hope this helps.

Thanks and regards,

Kushagra

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If the equation for the angular displacement of a particle moving on a circular path is given by theta=2t3 + 0.5 where theta is in radians and t in seconds, then the average angular velocity of the particle after 2 seconds from its start is?

Rituraj Singh

3 Answers

Arun Kumar

Nidhi

Kushagra Madhukar

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the earth’s gravitational pull (its weight) but deviates slightly from this direction. Calculate the
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