Respuesta :
Explanation:
According to the law of conservation of energy ,
Potential energy = kinetic energy
[tex]mgh = \frac{1}{2} \times mv^{2} + \frac{1}{2} \times I \times \omega^{2}[/tex]
I = [tex]\frac{mr^{2}}{2}[/tex]
[tex]\omega = \frac{v}{r}[/tex]
[tex]mgh = [\frac{1}{2} \times mv^{2}] + [\frac{1}{2} \times (\frac{mr^{2}}{2}) \frac{v^{2}}{r^{2}}][/tex]
mgh = [tex][\frac{1}{2} \times mv^{2}] + [\frac{1}{4} \times mv^{2}][/tex]
[tex]g \times h = \frac{3}{4} \times v^{2}[/tex]
[tex]9.8 \times 4.2 = \frac{3}{4} \times v^{2}[/tex]
v = 7.4 m/s
thus, we can conclude that the translational speed of the cylinder when it leaves the incline is 7.4 m/s.
The translational speed of the solid cylinder when it leaves the incline is 7.4 m/s.
Given the following data:
- Mass = 2.72 kg.
- Radius = 0.083 m.
- Height =4.20 m.
- Angle = 88.7°.
- Acceleration due to gravity = 9.81 [tex]m/s^2[/tex].
The angular velocity is given by:
[tex]\omega = \frac{V}{r}[/tex]
Mathematically, the moment of inertia of a solid cylinder is given by this formula:
[tex]I=\frac{1}{2} mr^2[/tex]
How to calculate the translational speed.
In accordance with the law of conservation of energy, the potential energy possessed by the mass at the beginning is equal to the kinetic energy possessed by the mass after falling:
[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2+\frac{1}{2} I\omega^2\\\\mgh = \frac{1}{2} mv^2+\frac{1}{2} \times (\frac{mr^2}{2} )(\frac{v}{r} )^2\\\\mgh=\frac{mv^2}{2} +\frac{mv^2}{4} \\\\V=\sqrt{\frac{4gh}{3} } \\\\V=\sqrt{\frac{4\times 9.8 \times 4.20}{3} }[/tex]
V = 7.4 m/s.
Read more on inertia here: https://brainly.com/question/3406242
