Answer:
a) [tex]K_{rot} = 44.005\,J[/tex], b) [tex]v \approx 3.997\,\frac{m}{s}[/tex], c) [tex]K_{2} = 72.472\,J[/tex], d) [tex]v \approx 3.244\,\frac{m}{s}[/tex]
Explanation:
b) The kinetic energy for a rigid body is the sum of translational and rotational kinetic energy:
[tex]K = K_{rot} + K_{tr}[/tex]
[tex]K = \frac{1}{2}\cdot \left(I\cdot \omega^{2} + m\cdot v^{2}\right)[/tex]
[tex]K = \frac{1}{2}\cdot \left[I\cdot \left(\frac{v}{R} \right)^{2} +m\cdot v^{2}\right][/tex]
[tex]K = \frac{1}{2}\cdot \left[\frac{I}{R^{2}}+\frac{3\cdot I}{2\cdot R^{2}} \right] \cdot v^{2}[/tex]
[tex]K = \frac{5\cdot I}{4\cdot R^{2}}\cdot v^{2}[/tex]
The speed of the center of mass of the sphere at the initial position is:
[tex]v = \sqrt{\frac{4\cdot K\cdot R^{2}}{5\cdot I} }[/tex]
[tex]v = \sqrt{\frac{4\cdot (110\,J)\cdot (0.130\,m)^{2}}{5\cdot \left(0.0931\,kg\cdot m^{2} \right)} }[/tex]
[tex]v \approx 3.997\,\frac{m}{s}[/tex]
a) The kinetic energy associated with the rotation is:
[tex]K_{rot} = \frac{1}{2}\cdot \left(0.0931\,\frac{kg}{m^{2}}\right)\cdot \left(\frac{3.997\,\frac{m}{s} }{0.130\,m} \right)^{2}[/tex]
[tex]K_{rot} = 44.005\,J[/tex]
It is 40 % of the total initial kinetic energy.
c) The total final kinetic energy is:
[tex]K_{2} = 110\,J - \left[\frac{3\cdot \left(0.0931\,kg\cdot m^{2}\right)}{2\cdot (0.130\,m)^{2}} \right]\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.20\,m)\cdot \sin 22.7^{\textdegree}[/tex]
[tex]K_{2} = 72.472\,J[/tex]
d) The speed of its center of mass is:
[tex]v = \sqrt{\frac{4\cdot K\cdot R^{2}}{5\cdot I} }[/tex]
[tex]v = \sqrt{\frac{4\cdot (72.472\,J)\cdot (0.130\,m)^{2}}{5\cdot \left(0.0931\,kg\cdot m^{2} \right)} }[/tex]
[tex]v \approx 3.244\,\frac{m}{s}[/tex]
