Question 413: Centre of Mass and Rotational Motion

Question

A cylinder is rolling down on an inclined plane of inclination $$60^{\circ}$$. It's acceleration during rolling down will be $$\frac{x}{\sqrt{3}} m / s^2$$, where $$x=$$ ________ (use $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$$).

Accepted Answer

To determine the acceleration of a cylinder rolling down an inclined plane without slipping, we can use Newton's second law and the concept of rolling motion. For an inclined plane at an angle $$ 	heta $$, the component of gravitational acceleration along the plane is $$ g \sin 	heta $$. However, because the cylinder is rolling and not sliding, not all of this component accelerates the center of mass; some of it goes into causing rotational acceleration about the center of mass.
For a rolling cylinder, the moment of inertia $$ I $$ is $$ I = \frac{1}{2} m r^2 $$, where $$ m $$ is the mass of the cylinder and $$ r $$ is the radius. The condition for rolling without slipping is that the linear acceleration $$ a $$ of the center of mass is equal to the radius $$ r $$ times the angular acceleration $$ \alpha $$, i.e., $$ a = r \alpha $$.
To find the linear acceleration $$ a $$, we use the torque $$ 	au $$ about the center of mass caused by the gravitational force down the incline. The torque due to gravity is $$ 	au = mg \sin 	heta \cdot r $$, and from Newton's second law for rotation, the angular acceleration is given by
$$ \alpha = \frac{	au}{I} = \frac{mg \sin 	heta \cdot r}{\frac{1}{2} m r^2} = \frac{2g \sin 	heta}{r} $$
Now using $$ a = r \alpha $$:
$$ a = r \left( \frac{2g \sin 	heta}{r} ight) = 2g \sin 	heta $$
But we must account for the fact that only a portion of the gravitational acceleration goes into translating the cylinder down the plane due to the rolling condition. This is where we apply the concept of the ``rolling factor'' for a cylinder, which is $$ \frac{2}{3} $$ for a solid cylinder, meaning $$ \frac{2}{3} $$ of the gravitational component is used for translation.
The acceleration of the center of mass for the cylinder is therefore:
$$ a = \frac{2}{3} g \sin 	heta $$
Now we plug in the values of $$ 	heta = 60^{\circ} $$ (which has $$ \sin 60^{\circ} = \frac{\sqrt{3}}{2} $$) and $$ g = 10 \; m/s^2 $$:
$$ a = \frac{2}{3} \cdot 10 \cdot \frac{\sqrt{3}}{2} = \frac{10 \sqrt{3}}{3} $$
To match the given expression $$ \frac{x}{\sqrt{3}} m / s^2 $$, let's manipulate our expression for $$ a $$:
$$ a = \frac{10 \sqrt{3}}{3} = \frac{10 \sqrt{3}}{3} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{10 \cdot 3}{3 \cdot \sqrt{3}} = \frac{10}{\sqrt{3}} m/s^2 $$
Hence, the value of $$ x $$ is $$ 10 $$.

Centre of Mass and Rotational Motion

A cylinder is rolling down on an inclined plane of inclination $60^{\circ}. It's acceleration during rolling down will be \frac{x}{\sqrt{3}} m / s^2, where x= ________ (use \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$).

Choose the correct answer: