A particle of mass m is released from point A as shown in the gigure. Would it loop the loop if H=4r? What is the force on the circular track when it is at point (a) B, (b) C?

Explanation

Let speed at the lowest point B ve `v_(B)`
Total energy A= Total energy at B
`0+mgH=1/2mv_(B)^(2)`
`v_(B)=sqrt(2gH)=sqrt(8gr)`
Let `N_(1)` be the force exerted by the track at B.
`N_(1)=mg+(mv_(B)^(2))/(R )`
If `v_(c)` is velocity at C.
`v_(C)^(2)=v_(B)^(2)-4gr`
`N_(2)` is force exerted by the track at C
`N_(2)=(mv_(c)^(2))/(r)-mg`
Let speed at lowest point B be `v_(B).`
Total energy at A= Total energy at B
`0+mgH=1/2mv_(B)^(2)+0`
`v_(B)=sqrt(2gh)=sqrt(2gxx4r)=sqrt(8gr)gtv_(min)(=sqrt(5gr))`
So, the particle can loop the loop
Let `N_(1)` be the force exerted by the track at B.

According to III law,
`N_(1)=mg+(mv_(B)^(2))/(r)`
`=mg+(mxx8gr)/(r)=9mg`

Force exerted by the particle on the track =9 mgh directed vertically downward At C,
Let the speed be `v_(C)`
`v_(C)^(2)=v_(B)^(2)-4gr`
`=8gr-4gr=4gr`
Let `N_(2)` be the force exerted by the track at C
`N_(2)=(mv_(C)^(2))/(r)-mg`
`=4mg-mg`
`=3mg`
Force exerted by the particle on the track at C is 3 mg directed redially outward.