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Shown a smooth looping the loop track A ...

Shown a smooth looping the loop track A particle of mass m is released from oint A , as shown If H = 3 r , whould the particle loop the loop ? What is the force on the circular track when the particle is at point (i) B (ii) C ? .

Text Solution

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Here , ` H = 3 r `
When the particle of mass m is released from A , velocity acquired by it on reaching B is
` upsilon_(B) = sqrt(2 g H ) = sqrt( 2 g xx 3 r ) = sqrt ( 6 gr) `
which is greater than sqrt(5 g r ) , the minimum velocity required at B for looping the loop Hence the particle will loop the loop
Force exerted by the circular track on the particle at B
`=mg +(mv_(B)^(2))/(r)`
`N_(1)=mg+ (m xx 6 gr)/(r)=7 mg`
Force exerted by circular track on the particle at C
`N_(2) = (m upsilon_(c)^(2))/(r) - mg , `
where ` upsilon_(C)^(2) = upsilon_(B)^(2)- 4 gr = 6 gr - 4 gr = 2 gr `
`:. N_(2) = (mxx2 g r )/(r) - mg = mg `
` N_(1) N_(2) ` are force exerted on the circular track by the particle at B , C respectively
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