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

D

Explanation

From F.B.D of ring, writing equation of motion of rir-
`2-f=2[0.3] rArr f = 2-0.6 = 1.4 `N……….(i)
As disc is rolling, `A = R alpha rArr 0.3 = alpha[0.5]`
`alpha =3/5 rad//s`……..(ii)
Now torque equation, `tau_( c) = I_( c)alpha`
`fR -2 muR = mR^(2)alpha, f-2mu = mR alpha, 1.4 - 2mu =2/3(3/5)`,
`=0.8 = 2mu rArr mu = 0.4 =P/10, therefore P=4`.