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A stepped cylinder having mass 50 kg and a radius of gyration (K) of 0.30 m. The radii R₁ and R₂ are respectively 0.3 m and 0.6 m. A pull T equal to 200 N is exerted on the rope attached to inner cylinder. The coefficient of static and dynamic friction between cylinder and ground are 0.1 and 0.08 respectively.

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The angular acceleration is 2.67 rad s^-2.

The force of kinetic friction is 40 N.

The acceleration is 3.2 m s^-2 towards left.

The acceleration is 3.2 m s^-2 towards right.

answer is A, B, C.

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Detailed Solution

If there is no slipping, then $τ = Iα$

$T [R_{2} - R_{1}] = [{MK}^{2} + {MR}_{2}^{2}] α$

$α = \frac{200 (0 .6 - 0 .3)}{50 (0 {.3}^{2} + 0 {.6}^{2})} = \frac{8}{3} = 2 .67 rad / s^{2}$

Using Newton's IInd law

$T - f = {MR}_{2} α$

$f = (T - {MR}_{2} α) = [200 - 50 \times 0 .6 \times \frac{8}{3}] = 120 N$

Now $f_{max} = μ_{s} mg = 0 .1 \times 50 \times 10 = 50 N$

$f_{\max} < f$ , hence slipping happens

$f_{k} = μ_{k} Mg = 0 .08 \times 50 \times 10 = 40 N$

$Ma = - T + 40 \Rightarrow a = - 3 .2 m / s^{2}$ ( towards left )

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