Answer:
[tex]K = 60.861\,J[/tex]
Explanation:
The moment of inertia of the merry-go-round is:
[tex]I = \frac{1}{2}\cdot \left(\frac{861\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot (1.84\,m)^{2}[/tex]
[tex]I = 148.618\,kg\cdot m^{2}[/tex]
The tangential acceleration experimented by the merry-go-round is:
[tex]a_{t} = \frac{57.6\,N}{\left(\frac{861\,N}{9.807\,\frac{m}{s} }\right) }[/tex]
[tex]a_{t} = 0.656\,\frac{m}{s^{2}}[/tex]
The linear speed after 2.54 seconds is determined hereafter:
[tex]v = 0\,\frac{m}{s} + (0.656\,\frac{m}{s^{2}} )\cdot (2.54\,s)[/tex]
[tex]v = 1.666\,\frac{m}{s}[/tex]
The angular speed of the merry-go-round is:
[tex]\omega = \frac{1.666\,\frac{m}{s} }{1.84\,m}[/tex]
[tex]\omega = 0.905\,\frac{rad}{s}[/tex]
The kinetic energy due to the rotation for the merry-go-round is:
[tex]K =\frac{1}{2}\cdot (148.618\,kg\cdot m^{2})\cdot (0.905\,\frac{rad}{s} )^{2}[/tex]
[tex]K = 60.861\,J[/tex]
