A disc shaped body has two tight windings of light threads - one on the inner rim of radius R = 1m and the other on outer rim of radius 2R (see figure). It is kept on a horizontal surface and the ends of the two threads are pulled horizontally in opposite directions with force of equal magnitude F = 20N.Mass of the body and its moment of inertia about an axis through centre O and perpendicular to the plane of the figure are $M = 4 kg and I = 8 kg - m^{2}$ respectively. Find the kinetic energy of the body 2 seconds after the forces begin to act, if
(i) the surface is smooth, (ii) the surface is rough enough to ensure rolling without sliding

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80 J , 38.28 J

56 J, 24.45J

100 J, 33.33 J

100J, 28.44 J

answer is C.

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Detailed Solution

(i) Net force on the body = 0
However, it has a net torque acting on it
$τ = F . 2 R - FR = FR = 20 \times 1 = 20 N - m$
$∴ α = \frac{τ}{I} = \frac{20}{8} = \frac{5}{2} rad / s^{2}$
Angular speed at t = 2s is w = at = 5 rad/s

$∴ K = \frac{1}{2} I (i)^{2} = \frac{1}{2} \times 8 \times 5^{2} = 100 . J$
(ii) Let friction be = f

$f = Ma \Rightarrow f = 4 a . . . (i) And Iα = F . 2 R - FR - f . 2 R \Rightarrow 8 . α = 20 \times 1 - 2 f \Rightarrow 4 α = 10 - f But 2 R \cdot α = a \Rightarrow 2 α = a ∴ 10 - f = 2 a$
Solving (i) and (ii) $a = \frac{10}{6} m / s^{2} = \frac{5}{3} m / s^{2}$

$α = \frac{a}{2 R} = \frac{5}{6} rad / s^{2}$
Hence $t = 2 s; V = at = \frac{10}{3} m / s and ω = αt = \frac{5}{3} rad / s$
$∴ K = \frac{1}{2} M V^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} \times 4 \times {(\frac{10}{3})}^{2} + \frac{1}{2} \times 8 \times {(\frac{5}{3})}^{2} = 33 .33 J$

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