Answer:
[tex]K.E=273.5J[/tex]
Explanation:
Given data
[tex]F_{g}=791N\\F_{h}=49N\\r=1.56m\\t=3.03s[/tex]
To find
Kinetic Energy
Solution
The moment of inertia is given as:
[tex]I=(\frac{1}{2} )MR^2\\I=\frac{1}{2}(F_{g}/g)R^2\\ I=\frac{1}{2}(\frac{791N}{9.8m/s^2} )(1.56m)^2\\ I=98.21kg.m^2[/tex]
The angular acceleration is given as:
[tex]\alpha =\frac{T}{I}\\\alpha =\frac{F_{y}R}{I}\\ \alpha =\frac{(49N)(1.56m)}{98.2kg.m^2}\\\alpha =0.778rad/s^2[/tex]
Now the angular velocity is given by:
[tex]w=\alpha t\\w=(0.778rad/s^2)(3.03s)\\w=2.36rad/s[/tex]
So the kinetic energy given as:
[tex]K.E=(\frac{1}{2} )Iw^2\\K.E=\frac{1}{2}(98.21kg.m^2)(2.36rad/s)^2\\ K.E=273.5J[/tex]
