IIT JEE , AIEEE Forum - Physics , Mechanics

newarnav

whats the prob buddy?

have you seen the Q i;ve mentioned in the earlier post

suresan

I think for A it is 2r and for B it is sqrt2 X r.

karthik2007

Your second part is pretty easy. The velocity of that point is the vector sum of two vectors perpendicular to each other, having a magnitude of r

each. Still working on the first part.

raulrag009

It's solution is quite sophisticated.

I took a bit of help from my maths teacher for this , even i could not get a bit of it , anywayz iam postin the solution

(1)The instantaneous centre of rotation for the rolling is the point of contact

with the surface.

(2)The trajectory traced out by each of the points on the suface of the cylinder

is a cycloid {as someone already mentioned}.

Consider it's parametric eqn's and solve .

Let $\phi$ be the angle turned through on generating circle by a point.

Radius of curvatur after solving , R = $4r{\sin{\frac{\phi}{2}}}$

Thus for point A $\phi$ = $\pi$ Hence radius of curvature ,R = $2\sqrt{2}(\sqrt{2})r=4r$

for point B $\phi$ = $\frac{3\pi}{2}$ R = $2\sqrt{2}R$

karthik2007

Could you mention its parametric equations? I know its something like x = t-sint, y = tcost... something like that

raulrag009

Here are the eqn's sir

$x=r(\phi-\sin{\phi})\\\\ y=r(1-\cos{\phi})$

karthik2007

Thanks Sir :)

Prakriteesh

No, the correct answers are 2r and root2 r. U see, the motion of the points is not cycloid if it is viewed from the frame fixed with the instantaneous axis (the axis passing thru the contact point). In that frame, the entire wheel rotates abt the instantaneous axis with the same angular velocity w, with the particles going in a purely circular manner. The distances of the points A and B from this axis are 2r and root2 r respectively. Hence these are the required radii of curvature (as the contact point itself is the centre of the circles in which the particles move). This inst. axis is a real axis as it remains fixed for a particular instant. So U need not add the linear motion of the cylinder.

Conjurer

^^

I agree completely.It will be a cycloid if seen from the ground frame only but if you are talking about radius of curvature its always performing circular motion about the instantaneous axis of rotation and I dont have any doubt that it should be 2r.

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