A disc is rotating as per the relation teta= at^2 + bt + c where a , b ,and c are constant t is the time and teta=angular displacement it is brought to rest in 10 rotations then b=?

Question

anonymous · Answer

teta?

anonymous · Answer

oh

anonymous · Answer

$$\theta=angular displacement$$

anonymous · Answer

What is the question?

anonymous · Answer

A disc is rotating as per the relation teta= at^2 + bt + c where a , b ,and c are constant t is the time and teta=angular displacement it is brought to rest in 10 rotations then b=?

anonymous · Answer

What is the answer?

anonymous · Answer

$$\sqrt{-80apie}$$

anonymous · Answer

@Ishaan94  wat do u get

anonymous · Answer

At t=0, the disc is already rotating. So, theta = c. (theta for teta)
After ten rotations theta=10*2pi. => 20pi = at^2 + bt + c. t is the time after 10 rotations. 

Is this all information you got? I am bit confused.

anonymous · Answer

Ohh theta is angular displacement. Sorry. I took it as the angular distance.

anonymous · Answer

I am still confused though.

anonymous · Answer

No Problem Bro just skip this and take a look at my New question plzz

anonymous · Answer

http://openstudy.com/study?signup#/updates/5017eeb5e4b04dfc808bb933

anonymous · Answer

$$\theta(t)=at^2+bt+c$$$$\omega(t)=d \theta(t)/dt=2at+b$$
when it stops:
$$\omega(T)=0=2aT+b \rightarrow T=-b/2a$$
Replacing in the angle equation:
$$\theta(T)=a(-b/2a)^2+b(-b/2a)+c=c-b^2/4a$$
$$\Delta \theta=\theta(T)-\theta(0)=c-b^2/4a-c=-b^2/4a=20\pi$$
$$b=\sqrt{-80\pi a}$$