Answer:
C. Both reach the bottom at the same time.
Explanation:
For a rolling object down an inclined plane , the acceleration is given below
a = g sinθ / (1 + k² / r² )
θ is angle of inclination , k is radius of gyration , r is radius of the cylinder
For cylindrical object
k² / r² = 1/2
acceleration = g sinθ /( 1 + 1/2 )
= 2 g sinθ / 3
Since it does not depend upon either mass or radius , acceleration of both the cylinder will be equal . Hence they will reach the bottom simultaneously.
The small light and large heavy cylinders will reach the bottom at the same time.
Let the mass of the small cylinder = m₁
Let the mass of the heavy cylinder = m₂
Apply the principle of conservation of mechanical energy;
The potential energy at the top = kinetic energy of each ball at the bottom
P.E = K.E
[tex]mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2} v^2\\\\h = \frac{v^2}{2g}[/tex]
where;
The height traveled by the cylinders is affected by their masses. If both cylinders traveled the same height, the time of motion will be the same.
Thus, we can conclude that both cylinders will reach the bottom at the same time.
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