Question

If the time of descent of a rolling cylinder on a plane inclined 100 to the horizontal is 3.35 seconds. What is the distance covered by the rolling cylinder, if the acceleration of the cylinder is given as (2gSinθ/3)? [Take g = 10 m/s2].

Answer 1

Given that,

Inclined angle = 10°

Time = 3.35 sec

If the acceleration of the cylinder is $2g\sin\theta=3a$

We need to calculate the acceleration

Using given acceleration

$2g\sin\theta=3a$

$a=\dfrac{2}{3}g\sin\theta$

Put the value into the formula

$a=\dfrac{2}{3}\times10\sin10$

$a= 1.15\ m/s^2$

We need to calculate the distance

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times1.15\times(3.35)^2$

$s=6.45\ m$

Hence, The distance covered by the rolling cylinder is 6.45 m.