Answer to Question #109472 in Mechanics | Relativity for let me know

Question

[img]https://upload.cc/i1/2020/04/13/d9J0an.jpg[/img]

Answer 1

As per the given question,

Mass of the block A(m)=4kg

Mass of the cylinder (M)=6kg

Radius of the cylinder (R)=20cm=0.2m

Let the tension in the string is T and the acceleration of the block is a, cylinder is rolling without slipping.

$mg-T=ma$

$4g-T=4a-----(i)$

Now, taking torque about the point of contact,

$I\alpha = TR$

$T=\dfrac{I\alpha}{R}=(MR^2+\dfrac{MR^2}{2})\dfrac{\alpha}{R}=\dfrac{3MR^2\times \alpha}{2R}=\dfrac{3Ma}{2}----(ii)$

From equation (i) and (ii)

$\Rightarrow 4g+\dfrac{3Ma}{2}=4a$

$\Rightarrow 4a+\dfrac{3\times 6a}{2}=4g$

$\Rightarrow 4a+9a=4g$

$\Rightarrow 13a=4g$

$\Rightarrow a=\dfrac{4g}{13}=3.01 m/sec^2$

Tension in the string, $T=\dfrac{3Ma}{2}=\dfrac{3\times 6\times 3.01}{2}27.09N$

ii) Speed of the cylinder after t=5 sec

$v=u+at$

$v=0+3.01\times 5= 15.05 m/sec$

angular velocity at t=5,

$\omega =\dfrac{v}{R}=\dfrac{15.05}{0.2}=75.25 rev/sec$

Hence kinetic energy of the cylinder =$\dfrac{MR^2}{2}+\dfrac{I\omega^2}{2}=\dfrac{6\times 15.05^2}{2}+\dfrac{3\times6\times 0.2^2\times 75.25^2}{2}$

$=679.50 +2038.52=2718.02J$