A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom

The correct answer is $\sqrt{\frac{4gh}{3}}$.

To find the speed of the centre of mass of the solid cylinder when it reaches the bottom of the inclined plane, we can apply the law of conservation of mechanical energy. Since the cylinder rolls without slipping, its mechanical energy is conserved throughout the motion.

Step 1: Initial potential energy at the top
Initially, the cylinder is at rest at height $h$. Its initial energy is purely gravitational potential energy:
$E_i=Mgh$
where $M$ is the mass of the cylinder, and $g$ is the acceleration due to gravity.

Step 2: Final kinetic energy at the bottom
When the cylinder reaches the bottom of the incline, its potential energy is converted entirely into kinetic energy. Since it is rolling without slipping, its kinetic energy is the sum of its translational kinetic energy and its rotational kinetic energy:
$E_f=K_t+K_r$
$E_f=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$
where:
- $v$ is the linear speed of the centre of mass,
- $I$ is the moment of inertia of the solid cylinder about its central axis, given by $I=\frac{1}{2}M{R}^2$, and
- $\omega$ is the angular velocity.

Step 3: Relationship between linear and angular velocities
Because the cylinder rolls without slipping, the relationship between its linear velocity and angular velocity is:
$\omega =\frac{v}{R}$

Step 4: Substitute and simplify the final kinetic energy
Substituting the expressions for $I$ and $\omega$ into the energy equation yields:
$E_f=\frac{1}{2}M{v}^2+\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\left(\frac{v}{R}\right)}^2$
$E_f=\frac{1}{2}M{v}^2+\frac{1}{4}M{v}^2$
$E_f=\frac{3}{4}M{v}^2$

Step 5: Equate initial and final energy
By conservation of energy, we set the initial energy equal to the final energy:
$Mgh=\frac{3}{4}M{v}^2$
Dividing both sides by $M$ gives:
$gh=\frac{3}{4}{v}^2$
Solving for $v$:
$v^2=\frac{4gh}{3}$
$v=\sqrt{\frac{4gh}{3}}$