IIT JEE , AIEEE Forum - Physics , Mechanics

sagar90

A PARTICLE IS MOVING ALONG A CIRCULAR PATH WITH ANGULAR SPEED

ABOUT D AXIS PASSING THRO ITS CENTRE WHAT WILL BE

ABOUT A POINT ON THE OTHER END OF DIAMETER THRO INSTANTANEOUS POSITION OF D PARTICLE????

pottermania1990

is it 2omega

sagar90

par kaise aya

sagar90

common yaar give d solution

rahulr

i think its omega/2 ...not 2*omega

sagar90

LET THE ANSWER BE ANYTHING BUT UR SOLN IS MUST YAAR

hardcore_genius

First of all hello Sagar, me also Sagar !!!

Now, i m giving only the hint that will take u 2 the solution.

Just make a circle & take any 2 points on its circumference.Now take a third point also on circumference.Join these 2 points with center (seperatlely) & then with third point.

Now, the angle subtended on the centre will double of that subtended on the third point. (basic theorem). Let, the angle on centre be ? & on third point be @ respectively.

Now, ? = @ *2

differentiating wrt time, u will get,

angular velocity wrt center = 2 * angular velocity wrt any point on the circumference.

even if u have any doubt, then don't hesitate to nudge me.

hsbhatt

= v_perp/r.

Here v_perp = r_o

and r = 2r_o. Hence,

(on circ) =

/2.

The previous solution is good for this prob. Now try solving this prob:

Choose a point on a disc rotating with angular speed

an axis perp to the plane of the disc and passing thru its centre. What is it's angular vel with respect to the point on the opposite end of the diameter. And w.r.t a point at an angle

to the given point.

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