1Got it sir!!
Using energy conservation, the answer comes out to be √8gr.
The maximum possibility for the ring to bounce off is when the body is at the topmost position. Hence we will equate centripetal force to weight.
A small body 'A' is fixed to the inside of a thin rigid hoop of radius R and mass equal to that of body 'A'. The hoop rolls without slipping over a horizontal plane. At the instant when the body is at lowermost position, the centre of the hoop moves with velocity v. For what values of v will the hoop move without bouncing?
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12 Answers
19use the concepts of rolling, along with the fact that the question is trying to say :" normal force of the small object never becomes zero" !
62Draw the fbd at an angle theta..
Find the relation between the velocities... at an angle theta... (by energy conservation?)
Now try?
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66@iota.1
can you show how it is coming to be √8gR by energy conservation?
3N-mgcosθ=mva2/R
N≥0 and va2=2gR(1-cosθ)
mg(3-cosθ)≥0 minimum value is cosθ=1/3
minimum velocity is √4/3gR
66@msp
that's definitely not correct. There are quite a few mistakes. First specify the reference frame. Because w.r.t the ground, the particle A is not moving in a circle. Secondly, how did you get va2=2gR(1-cosθ)?
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At this instant, 2mrcmω'2=2mg
Ei=12Ipω02
Ef=12I'pω'2+(2m)gr
I'p=6mr2
I got the answer using these equations.
