Answer:
K.E=365.2 J
Explanation:
Given data
Weight w =953 N
radius r=1.68 m
F=73.9 N
t=2.55 s
g=9.8 m/s²
To find
Kinetic Energy K.E
Solution
From the moment of inertia
[tex]I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}[/tex]
The angular acceleration is given as
[tex]a=T/I\\a=\frac{FR}{I}\\ a=\frac{(73.9)(1.68)}{137.232}\\a=0.905rad/s^{2}[/tex]
The angular velocity is given as
[tex]w=at\\w=(0.905)(2.55)\\w=2.31rad/s[/tex]
So the Kinetic Energy is given as
[tex]K.E=(1/2)Iw^{2}\\ K.E=(1/2)(137.232)(2.31)^{2}\\ K.E=365.2J[/tex]
