Answer:
The rotational inertia of my wheel is [tex]I =1.083 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The mass of the wheel is [tex]m = 1.6 \ kg[/tex]
The radius of the wheel is [tex]r = 0.37 \ m[/tex]
The height is [tex]h = 6.7 m[/tex]
The linear speed is [tex]v = 4.7 m/s[/tex]
According to the law of energy conservation
[tex]PE = KE + KE_R[/tex]
Where PE is the potential energy at the height h which is mathematically represented as
[tex]PE = mgh[/tex]
While KE is the kinetic energy at the bottom of height h
[tex]KE = \frac{1}{2} mv^2[/tex]
Where [tex]KE_R[/tex] is the rotational kinetic energy which is mathematically represented as
[tex]KE_R = \frac{1}{2} * I * \frac{v^2}{r^2}[/tex]
Where [tex]I[/tex] is the rotational inertia
So substituting this formula into the equation of energy conservation
[tex]mgh = \frac{1}{2} mv^2 + \frac{1}{2} * I * \frac{v^2}{r^2}[/tex]
=> [tex]I =[ \ mgh - \frac{1}{2} mv^2 \ ]* \frac{2 r^2}{v^2}[/tex]
substituting values
[tex]I =[ \ 1.6 * 9.8 * 6.7 - \frac{1}{2} * 1.6 *4.7^2 \ ]* \frac{2 * 0.37^2}{4.7^2}[/tex]
[tex]I =1.083 \ kg \cdot m^2[/tex]
