Question 124: Centre of Mass and Rotational Motion

Question

A uniform solid cylinder of mass ' m ' and radius ' r ' rolls along an inclined rough plane of inclination $45^{\circ}$. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be

Accepted Answer

The acceleration of the center of mass for a rolling object down an incline is given by:
$$ a = \frac{g \sin	heta}{1 + \frac{I}{m r^2}} $$
For a uniform solid cylinder, the moment of inertia about its central axis is:
$$ I = \frac{1}{2} m r^2 $$
Substituting this into the expression for acceleration:
$$ a = \frac{g \sin	heta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin	heta}{1 + \frac{1}{2}} = \frac{g \sin	heta}{\frac{3}{2}} = \frac{2}{3} g \sin	heta $$
For an incline of $$45^\circ$$, we have:
$$ \sin 45^\circ = \frac{\sqrt{2}}{2} $$
Thus, the acceleration becomes:
$$ a = \frac{2}{3} g \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} g}{3} $$
Therefore, the linear acceleration of the cylinder's axis is:
$$ \frac{\sqrt{2} g}{3} $$

Answer

$\sqrt{2} \mathrm{~g}$

Answer

$\frac{1}{\sqrt{2}} \mathrm{~g}$

Answer

$\frac{1}{3 \sqrt{2}} \mathrm{~g}$

Answer

$\frac{\sqrt{2} g}{3}$

Centre of Mass and Rotational Motion

A uniform solid cylinder of mass ' m ' and radius ' r ' rolls along an inclined rough plane of inclination 45^{\circ}. If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder's axis will be

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