A solid brass cylinder and a solid wood cylinder have the same radius and mass (the wood cylinder is longer). Released together from rest, they roll down an incline. (a) Which cylinder reaches the bottom first, or do they tie? (b) The wood cylinder is then shortened to match the length of the brass cylinder, and the brass cylinder is drilled out along its long (central) axis to match the mass of the wood cylinder.Which cylinder now wins the race, or do they tie?

Both wood and brass cylinders have the same radius and mass and initially, both are at rest before rolling down an incline.

The torque acting on the object is equal to the moment of force. The toque is also equal to the product of the moment of inertia and angular acceleration of the object. It can be written as the time rate of change of angular momentum of the object.The moment of inertia of an object is equal to the sum of the product of mass and the perpendicular distance of all the points from the axis of rotation.

Using the equations of the moment of inertia of the cylinder, torque, and relation between angular acceleration and linear acceleration, find which cylinder will reach the bottom first.

Formulae are as follows:

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

$x=v_{0}t+\frac{1}{2}a{t}^2$

$\alpha =\frac{a}{R}$

$\tau =I\alpha$

whereL is angular momentum,$\tau$ is torque,m is mass, R is a radius, a is acceleration,$\alpha$is angular acceleration, t is time, x is displacement,I is a moment of inertia andvis velocity.

Now,

${\tau }_{net}=I\alpha$

For cylinder,

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

$I=\frac{3}{2}m{R}^2$

And, $\alpha =\frac{a}{R}$,

${\tau }_{net}=\frac{3}{2}m{R}^2\times \frac{a}{R}$

But,

${\tau }_{net}=mg\sin \theta R$

Therefore,

$mg\sin \theta R=\frac{3}{2}m{R}^2\times \frac{a}{R}$

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

Now,

$x=v_{0}t+\frac{1}{2}a{t}^2$

As,$v_0=0\text{m/s}$,

$t=\sqrt{\frac{2x}{a_{}}}$

Here, it doesn't depend on the mass and radius, therefore both will tie.

Now, for the brass cylinder,

$I_1=\frac{1}{2}m(R_1^2+R_2^2)$

But,

${\tau }_{net}=I_1\alpha$

And,$\alpha =\frac{a_1}{R}$,

${\tau }_{net}=I_1\frac{a_1}{R}$

But,

${\tau }_{net}=mg\sin \theta R$

Therefore,

$mg\sin \theta R=I_1\frac{a_1}{R}$

$a_1=\frac{mg\sin \theta R^2}{\frac{1}{2}m(R_1^2+R_2^2)}$

For wood cylinder,

$I_2=\frac{1}{2}m{R}_2^2$

But,

${\tau }_{net}=I_2\alpha$

And,

$\alpha =\frac{a_2}{R}$

${\tau }_{net}=I_2\frac{a_2}{R}$

${\tau }_{net}=I_2\frac{a_2}{R}$

But,

${\tau }_{net}=mg\sin \theta R$

Therefore,

$\begin{array}{c}mg\sin \theta R=I_2\frac{a_2}{R}\\ a_2=\frac{mg\text{sin}\theta R^2}{I_2}\\ =\frac{mg\text{sin}\theta R^2}{\frac{1}{2}m{R}_2^2}\end{array}$

As,

$I_1>I_2$

Therefore,.$a_2<a_1$

Hence, wood cylinder will win.

Therefore, using the equations of moment of inertia of the cylinder, torque and relation between angular acceleration and linear acceleration, which cylinder will reach the bottom first can be found.