Respuesta :
Answer:
The kinetic energy of the merry-go-round is [tex]\bf{475.47~J}[/tex].
Explanation:
Given:
Weight of the merry-go-round, [tex]W_{g} = 826~N[/tex]
Radius of the merry-go-round, [tex]r = 1.17~m[/tex]
the force on the merry-go-round, [tex]F = 57.8~N[/tex]
Acceleration due to gravity, [tex]g= 9.8~m.s^{-2}[/tex]
Time given, [tex]t=3.47~s[/tex]
Mass of the merry-go-round is given by
[tex]m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg[/tex]
Moment of inertial of the merry-go-round is given by
[tex]I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}[/tex]
Torque on the merry-go-round is given by
[tex]\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m[/tex]
The angular acceleration is given by
[tex]\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}[/tex]
The angular velocity is given by
[tex]\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}[/tex]
The kinetic energy of the merry-go-round is given by
[tex]E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J[/tex]
