Sliding rod (II)

A thin rod of mass m m and length l l is kept vertically on a smooth table. It is given a slight push such that it starts moving due to uniform gravitational field g g present in vertically downwards direction. The acceleration of center of mass is given by a c m ( θ ) = g a ( b a s i n 2 θ ) c ( d + e s i n θ f s i n 2 θ + i s i n 4 θ ) { a }_{ cm }(\theta )\quad =\quad g\cdot \frac { a }{ { (b-a{ sin }^{ 2 }\theta ) }^{ c } } \left( d+esin\theta -f{ sin }^{ 2 }\theta +i{ sin }^{ 4 }\theta \right) where a a , b b , c c , d d , e e , f f and i i are positive integers. Find the value of a + b + c + d + e + f + i a+b+c+d+e+f+i .


The answer is 27.

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1 solution

Ranajay Medya
May 4, 2018

Equations represent:
1. Force equation
2. Torque equation
3. Constraint relation
I. Conservation of energy
II. Constraint relation


Calculations can easily be simplified using IAOR for the rod And considering pure rotation about IAOR

Suhas Sheikh - 2 years, 12 months ago

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