Answer :
The speed of the sphere's center of mass at the bottom of the incline is 2.79 m/s.
- Given data:
- Mass of the sphere, [tex]\( m = 3.80 \)[/tex] kg
- Height of the incline, [tex]\( h = 1.55 \)[/tex] m
- Incline angle, [tex]\( h = 1.55 \)[/tex]
- To find the speed of the sphere's center of mass at the bottom of the incline, we can use the conservation of energy principle, considering both translational and rotational kinetic energies.
1. Potential Energy at the Top:
[tex]\[ PE_{\text{top}} = mgh \][/tex]
Where:
- [tex]\[ PE_{\text{top}} = mgh \][/tex] is the acceleration due to gravity [tex](\( 9.81 \, \text{m/s}^2 \))[/tex].
Substituting the given values:
[tex](\( 9.81 \, \text{m/s}^2 \))[/tex]
2. Kinetic Energy at the Bottom:
At the bottom of the incline, the energy will be in the form of translational kinetic energy and rotational kinetic energy (since it rolls without slipping).
- Translational kinetic energy:
[tex]\[ KE_{\text{trans}} = \frac{1}{2} mv^2 \][/tex]
- Rotational kinetic energy (for a sphere rolling without slipping):
[tex]\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \][/tex]
Where [tex]\( I \)[/tex] is the moment of inertia of the sphere about its center and [tex]\( \omega \)[/tex] is the angular velocity.
For a solid sphere:
[tex]\[ I = \frac{2}{5} mR^2 \][/tex]
Where [tex]\( R \)[/tex] is the radius of the sphere.
- The relationship between translational and rotational speeds for rolling without slipping:
[tex]\[ v = \omega R \][/tex]
3. Total Energy Conservation:
Since energy is conserved, we equate [tex]\( PE_{\text{top}} \) to \( KE_{\text{trans}} + KE_{\text{rot}} \)[/tex]:
[tex]\( PE_{\text{top}} \) to \( KE_{\text{trans}} + KE_{\text{rot}} \)[/tex] Substituting \( I = \frac{2}{5} mR^2 \):
[tex]\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \][/tex]
Simplifying and solving for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{\frac{10gh}{7}} \][/tex]
4. Calculate the Final Velocity:
Now substitute the known values:
[tex]\[ v = \sqrt{\frac{10 \times 9.81 \times 1.55}{7}} \approx 2.79 \, \text{m/s} \][/tex]
Complete Question:
A solid homogeneous sphere of mass [tex]\( m = 3.80 \)[/tex] kg is released from rest at the top of an incline of height [tex]\( h = 1.55 \)[/tex] m and rolls without slipping to the bottom. The incline is at an angle [tex]\( \theta = 27.7^\circ \)[/tex] to the horizontal. Calculate the speed of the sphere's center of mass at the bottom of the incline.
