Try this:
The rigid body is so selected from a collection of spheres, rings, discs etc. such that when released from rest in the position shown, it leaves contact with the horizontal plane at exactly 60*...as shown.
This would assist in a smooth 'transition' from the horizontal to the inclined surface;
So, whats the nature of the rigid body???
the moment of inertia about the instantaneous axis of rotation is 2MR^2.....?
ReplyDeleteyeahh...getting same as sambhav
ReplyDeleteits a ring
...it wasnt very difficult..coz we have a similar question in module
ring is true...
ReplyDeleteDerive the angle A rotated by a general rigid body of parameters M,K,R.
It is given by CosA=2/(3+k^2/R^2)...
For a particle: k^2/R^2=0 or CosA=2/3.
Rings a bell???
thats a shocking result.REASON?????????
ReplyDeletehere, the body rolls about a point (the corner made by the two surfaces)
ReplyDeleteactually, you can keep the body on top a fixed round surface of radius R (convex up), and let it roll down. it will leave contact at the same angle..ie CosA=2/(3+k^2/R^2)
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ReplyDeleteplease only explain the reason bhaiya that for a particle CosA=2/3.Mathematically i agree.but practically i dont.
ReplyDeleteyou agree for particle sliding on a fixed sphere??
ReplyDeleteThis is just a limiting case when the radius of the fixed sphere tends to zero.