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

Question #109472
1
2020-04-15T10:43:28-0400

As per the given question,

Mass of the block A(m)=4kg

Mass of the cylinder (M)=6kg

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"

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