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

If net force applied by the rod is considered to be 2N.
`sqrt(f.^(2)+F^(2)=2`........(i)
`FR-f.R=2mR^(2).(a)/(R)`
`F-f.=2ma=1.2`
From Eqs. (i) and (ii),
`(1.2+f.)^(2)+f.^(2)=2^(2)`.....(ii)
`2f.^(2)+2.4f.+1.44=4`
`f.^(2)+1.2f.+0.72-2=0`
`f.^(2)+1.2f.-1.28=0`
`f.=(-1.2+-sqrt(1.44+4xx(1.28)))/(2)`
`=0.6+-sqrt(0.36+1.28)=-0.6+-sqrt(0.64)=0.68`
From Eq. (ii), F=`1.88`.
`mu=(0.68)/(1.88)=(P)/(10)impliesP=3.16~~4`
But if only normal reaction applied by the rod is considered to be 2N.
`II^(nd)` Law `implies2-f=2[0.3]impliesf=2-0.6`
`f=1.4Nx`
`A=R alphaimplies0.3=alpha[0.5]impliesalpha=(3)/(5)rad//s`
`tau_(c)=I_(c)alphaimpliesfR-2muR=mR^(2)alpha`
`f-2mu=mRalpha`
`1.4-2mu=(2)/(2)((3)/(2))implies1.4-0.6=2mu`
`0.8=2muimpliesmu0.4=(P)/(10) :. P=4`