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A ball of mass 'm' is released from poin...

A ball of mass `'m'` is released from point `A` where, `Theta_(0)=53^(@)` . Length of pendulum is `'l'` . Find `v,a_(r),a_(t),a` , `T` and `F_(n et)` at
(a) point `A`
(b) point `C`
(c) pont `P` where `theta=37^(@)`

Text Solution

Verified by Experts

(a) At pont `A`
`v=0 implies a_(r)=(v^(2))/(R)=0` `(R=l)`
`a_(t)=gsintheta_(0)=gsin53^(@)=(3)/(5)g`
`a=a_(t)=(4)/(5)mg`
`F_(n et)=ma=(4)/(5)mg`
(b) At point `C`
`h=OC-OM=l-lcos53^(@)`
`=l-(3)/(5)=(2)/(5)l`
`v=sqrt(2gh)=sqrt(2g((2)/(5)l))=sqrt((4)/(5)gl)`
`a_(r)=(v^(2))/(R)=((sqrt((4)/(5)gl)^(2)))/(l)=(4)/(5)g`
`a_(t)=0`
`a=a_(r)=(4)/(5)g`
`T-mg=(mv^(2))/(R)=(4)/(5)mg`
`T=(9)/(5)mg`
`F_(n et)=ma=(4)/(5)mg`

(c) At point `P`
`h=OM-ON-lcos37^(@)-lcos53^(@)`
`=(4)/(5)l-(3)/(5)l=(l)/(5)`
`v=sqrt(2gh)=sqrt(2g((l)/(5)))=sqrt((2)/(5)gl)`

`a_(r)=(v^(2))/(R)=((sqrt((2)/(5)gl)))/(l)=(2)/(5)g`
`a_(t)=gsintheta=gsin37^(@)=(3)/(5)g`
`a=sqrt(a_(r)^(2)+a_(t)^(2))=sqrt(((2)/(5)g)^(2)+((3)/(5)g)^(2))=(sqrt(13))/(5)g`
`T-mgcostheta=ma_(r)`
or `T-mgcos37^(@)=m((2)/(5)g)`
`T=(6)/(5)mg`
`F_(n et)=ma=(sqrt(13))/(5)mg`
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