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" !
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
19
62Draw the fbd at an angle theta..
Find the relation between the velocities... at an angle theta... (by energy conservation?)
Now try?
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.
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)?
1
At this instant, 2mrcmω'2=2mg
Ei=12Ipω02
Ef=12I'pω'2+(2m)gr
I'p=6mr2
I got the answer using these equations.
