Answer :
Answer:
a) [tex]\omega_{f} = 1.012\,\frac{rad}{s}[/tex], b) [tex]\Delta K = - 149.352\,J[/tex]
Explanation:
a) The initial angular speed of the merry-go-round is:
[tex]\omega_{o} = \left(\frac{1\,rev}{4\,s}\right)\cdot \left(\frac{2\pi\,rad}{1\,rev} \right)[/tex]
[tex]\omega_{o} \approx 1.571\,\frac{rad}{s}[/tex]
The final angular speed of the merry-go-round is computed with the help of the Principle of Angular Momentum:
[tex]\left(340\,\frac{kg}{m^{2}}\right)\cdot \left(1.571\,\frac{rad}{s} \right) = \left[340\,\frac{kg}{m^{2}}+(25\,kg)\cdot (2.74\,m)^{2} \right]\cdot \omega_{f}[/tex]
[tex]\omega_{f} = 1.012\,\frac{rad}{s}[/tex]
b) The change in kinetic energy is:
[tex]\Delta K = \frac{1}{2}\cdot \left\{\left[340\,\frac{kg}{m^{2}} + \left(25\,kg\right)\cdot \left(2.74\,m\right)^{2}\right]\cdot \left(1.012\,\frac{rad}{s} \right)^{2} - \left(340\,\frac{kg}{m^{2}} \right)\cdot \left(1.571\,\frac{rad}{s} \right)^{2} \right\}[/tex][tex]\Delta K = - 149.352\,J[/tex]