To solve this problem, we will use the principles of energy conservation.
### Given Data:
- Length of the inclined plane, \( s = 30 \) m
- Angle of inclination, \( \theta = 30^\circ \)
- Initial velocity, \( u = 0 \) (since the cylinder starts from rest)
- Acceleration due to gravity, \( g = 9.8 \) m/s²
### Step 1: Potential Energy at the Top
The height of the cylinder from the ground is given by:
\[ h = s \sin \theta \]
Substituting the given values:
\[ h = 30 \sin 30^\circ \]
\[ h = 30 \times 0.5 = 15 \text{ m} \]
The potential energy at the top is:
\[
PE = mgh = m \times 9.8 \times 15 = 147m \text{ J}
\]
(where \( m \) is the mass of the cylinder).
### Step 2: Kinetic Energy at the Bottom
When the cylinder reaches the bottom, all the potential energy converts into kinetic energy. Since the cylinder rolls without slipping, its total kinetic energy consists of both translational and rotational kinetic energy.
For a rolling body:
\[
KE_{\text{total}} = KE_{\text{translational}} + KE_{\text{rotational}}
\]
- Translational kinetic energy:
\[
KE_{\text{translational}} = \frac{1}{2} m v^2
\]
- Rotational kinetic energy:
\[
KE_{\text{rotational}} = \frac{1}{2} I \omega^2
\]
For a solid cylinder, the moment of inertia about its own axis is:
\[
I = \frac{1}{2} m R^2
\]
Since the cylinder rolls without slipping:
\[
v = R \omega
\]
Thus,
\[
\omega = \frac{v}{R}
\]
Substituting this in the rotational kinetic energy equation:
\[
KE_{\text{rotational}} = \frac{1}{2} \times \frac{1}{2} m R^2 \times \left(\frac{v}{R}\right)^2
\]
\[
KE_{\text{rotational}} = \frac{1}{4} m v^2
\]
### Step 3: Applying Energy Conservation
\[
mgh = KE_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{4} m v^2
\]
Factoring out \( m v^2 \):
\[
mgh = \left(\frac{1}{2} + \frac{1}{4}\right) m v^2
\]
\[
mgh = \frac{3}{4} m v^2
\]
Canceling \( m \) from both sides:
\[
gh = \frac{3}{4} v^2
\]
Substituting \( g = 9.8 \) m/s² and \( h = 15 \) m:
\[
(9.8)(15) = \frac{3}{4} v^2
\]
\[
147 = \frac{3}{4} v^2
\]
Multiplying by \( \frac{4}{3} \) on both sides:
\[
v^2 = \frac{4 \times 147}{3}
\]
\[
v^2 = \frac{588}{3} = 196
\]
\[
v = \sqrt{196} = 14 \text{ m/s}
\]
### Final Answer:
The velocity of the cylinder at the bottom of the inclined plane is **14 m/s**.
