Question

a cylinder is rolling down a n inclined plane of inclination 60o . What is its acceleration?
1) g/ root 3 2) g.root3 3) underoot 2g/3 4) none of these

Answer 1

the correct option is a

Answer 2

Q) a cylinder is rolling down an inclined plane of inclination of 60°. What is its acceleration?

1) $\frac{g}{\sqrt{3} }$           2) $g\sqrt{3}$         3)  $\sqrt{\frac{2g}{3} }$        4) None of these

Option (1) $\frac{g}{\sqrt{3} }$ is the answer.

Given,

A cylinder is rolling down an inclined plane of inclination of 60°.\

To Find,

Acceleration of the cylinder.

Solution,

The formula to find out the acceleration of the cylinder is,

a = ( g sinФ) / $1+\frac{K^{2} }{R^{2} }$

Here, the value of Ф is 60°.

In the case of a cylinder,

I = $\frac{1}{2}$m$R^{2}$

$\frac{1}{2} mR^{2}$ = m$K^{2}$

$K^{2}$ = $\frac{R^{2} }{2}$

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

Then,

a = $\frac{g sin 60}{1+\frac{1}{2} }$

a = $\frac{g \frac{\sqrt3} {2} }{\frac{3}{2} }$

After simplification, we get the answer,

a  = $\frac{g}{\sqrt{3} }$

Hence,  $\frac{g}{\sqrt{3} }$ is the acceleration of the rolling cylinder.

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