Answer:
71.42857 kgm²
1311.81946 kg
5.082 rad/s
Explanation:
F = Force = 200 N
R = Radius = 0.33 m
[tex]\alpha[/tex] = Angular acceleration is 0.924 rad/s²
Moment of inertia is given by
[tex]I=\dfrac{FR}{\alpha}\\\Rightarrow I=\dfrac{200\times 0.33}{0.924}\\\Rightarrow I=71.42857\ kgm^2[/tex]
The moment of inertia of the wheel is 71.42857 kgm²
Moment of inertia of a solid cylindrical wheel is given by
[tex]I=\dfrac{1}{2}mR^2\\\Rightarrow m=\dfrac{2I}{R^2}\\\Rightarrow m=\dfrac{2\times 71.42857}{0.33^2}\\\Rightarrow m=1311.81946\ kg[/tex]
The mass of the wheel is 1311.81946 kg
Rotational impulse is given by
[tex]I_r=\tau t\\\Rightarrow I_r=FRt[/tex]
Also
[tex]I_r=I(\omega_f-\omega_i)\\\Rightarrow I_r=I(\omega_f-0)\\\Rightarrow I_r=I\omega_f\\\Rightarrow \omega_f=\dfrac{I_r}{I}[/tex]
[tex]\omega_f=\dfrac{FRt}{I}\\\Rightarrow \omega_f=\dfrac{200\times 0.33\times 5.5}{71.42857}\\\Rightarrow \omega_f=5.082\ rad/s[/tex]
The angular speed is 5.082 rad/s