A boy is pushng a ring of mass 2kg and radius 0.5 m with a stick as shwon in figure. The stick applies a force of 2N on the ring and rolls it without slipping with an accelertaion of `0.3 m/s^2.` The coefficinet of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring of (P/10). The value of P is

Answer

4

Explanation

`sqrt(f^('2)+F^(2))=2`……….i
`FR-f'R=w2mR^(2)a/R`
`F-f'=2ma=1.2`………….ii
From eqn i and ii `(1.2+f')^(2)+f^('2)=2^(2)`
`2f^('2)+1.2f'+0.72-2=0`
`f^('2)+1.2f'-1.28=0`
`f'=(-1.2+-sqrt(1.44+4xx91.28))/2`
`=0.6+-sqrt(0.36+1.280=-0.6+-sqrt(0.64)=0.68`
From eqn ii `F=1.88.`
`mu=0.68/1.88=P/10impliesP=3.16~~4`
II law `implies 2-f=2[0.3]=.f=2-0.6`
`f=1.4Nx`.........i
`A=Ralpha`
`implies 0.3=alpha[0.5]` ,brgt `implies alpha=3/5rad//s`
`tau_(c)=I_(c)alphaimplies fR-2muR=mR^(2)alpha`
`f-2mu=mRalpha`
`1.4-2mu=2/2(3/2)`
`=1.4-0.6=2mu`
`0.8=2mimplies mu=0.4=P/10`
`:. P=4`