Question

A smooth track in the form of a quarter-circle of radius 6 m lies in the vertical plane. A ring of weight 4(N) moves from `P_(1)` and `P_(2)` under the action of forces `F_(1) , F_2` and `F_3`. is always towards `P_(2)` under the action of forces `F_(1),F_(2)` and `F_(3)` Force `F_(1)` is always towards `P_(2)` and is always (20) N in maghitude, force `F_(2)` always acts hotizontally and is always (30 N) in magnitude, force `F_(3)` always acts tangentially to the track and is of magnitude `(15-10 s) N`, where s is in metre. If the particle has speed `4 m//s` at `P_(1)`,what will its speed be at `P_(2)`?

Answer 1

The work done by `F(1)` is
`W_(1)=int_(P1)^(P2)F_(1) costheta ds`
From figure, `s=R(pi/2-2theta)`
or `ds=(6m)d(-2theta)=-12d theta`
and `F_(1)=20`.
At `P_(1)`, `2theta =pi/2 rArrtheta=pi/4`
At, `P_(2), 20=0 rArr theta=0`
Hence,` W_(1)=-240 int_(pi//4)^(0)cos theta d theta`
`=240 sinpi/4 =120sqrt2 J`

the work done by `F_(3)` is
`W_(3) =int F_(3)ds =int_(0)^(6(pi//2))(15-10s)ds , (P_(1)P_(2)=R(pi)/(2) =(6pi)/2)`
`[15s-5s^(2)]_(0)^(3pi) =-302.8J`
To calculate the work done by `F_(2)` and by w, it is convenient to take the projection of the path in the direction of the force. Thus,
`W_=(2)F_(2) (OP_(2)) =30(6) =180 J`
`W_(4)=(-w)(P_(1)O)=(-4)(6)=-24 J` (w=weight)
The total work done is , `W_(1) + W_(3) + W_(2) + W_(4) =23 J`
Then, by the work-energy principle.
`K_(P2)-K_(P1) =23 J`
`=1/2(4/(9.8))v_(2)^(2)-1/2(4/(9.8))(4)^(2) =23`.
`v_(2) = 11.3 m//s`.