Question

A thin circular ring first lives down a smooth incline then rolls down a rough incline of analytical geometry from same height ratio of time taken into motion is

Answer 1

The ratio of time taken into motion is 1 : 1.414

Explanation:

Let say the length of the inclined plane is L

so we will have time to slide it down is given as

$L = \frac{1}{2}at^2$

here we know that

$a = g sin\theta$

$L = \frac{1}{2}(gsin\theta} t^2$

$t_2 = \sqrt{\frac{2L}{gsin\theta}}$

now if the ring is rolling down the plane

the acceleration of the rolling ring is given as

$a = \frac{gsin\theta}{1 + \frac{k^2}{R^2}}$

for ring we know that

$\frac{k^2}{R^2} = 1$

$a = \frac{gsin\theta}{2}$

now time taken by the ring to roll down is given as

$t_2 = \sqrt{\frac{4L}{gsin\theta}}$

now ratio of the two time is given as

$t_1 : t_2$ = 1 : 1.414

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Topic : Rolling on inclined plane

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