Answer:
Ktot = 60.8 J
Explanation:
- The total kinetic energy of a rolling ball, is composed by a rotational kinetic energy (due to the rotation around a diameter), plus a traslational kinetic energy, due to the translation of the center of mass.
- The rotational kinetic energy can be written as follows:
[tex]K_{rot}= \frac{1}{2}* I * \omega^{2} (1)[/tex]
- where I = moment of Inertia of a solid sphere, and ω², the square of the angular speed.
- If the ball rolls without slipping, there exists a fixed relationship between the angular speed and the speed of the center of mass, as follows:
[tex]\omega = \frac{v_{CM} }{R} (2)[/tex]
- The traslational part, can be expressed as follows:
[tex]K_{trasl} = \frac{1}{2} * m * v_{CM} ^{2} (3)[/tex]
- For a solid sphere, like a bowling ball, the moment of inertia about an axis passing through a diameter can be written as follows:
[tex]I =\frac{2}{5} * m * R^{2} (4)[/tex]
- Replacing (2) and (4) in (1), we have:
[tex]K_{rot}= \frac{1}{2}* \frac{2}{5} * m * R^{2} * (\frac{v_{CM} }{R}) ^{2} = \frac{1}{5} * m* v_{CM} ^{2} (5)[/tex]
- The total kinetic energy is just the sum of (3) and (5):
[tex]K_{tot} = \frac{1}{5} * m* v_{CM} ^{2} + \frac{1}{2} * m * v_{CM} ^{2} = \frac{7}{10} * m* v_{CM} ^{2} (6)[/tex]
- The speed of the center of mass is given in km/h, so it is needed to convert it to m/s, as follows:
[tex]v_{CM} = 15.0 \frac{km}{h} * \frac{1000m}{1km} * \frac{1h}{3600s} = 4.17m/s (7)[/tex]
- Replacing (7) and m=5.00 kg in (6), we get the total kinetic energy of the rolling ball, as follows:
[tex]K_{tot} = \frac{7}{10} * 5.00 kg * (4.17 m/s) ^{2} = 60.8 J[/tex]
