Answer:
I = 70.86 kg.m²
Explanation:
given,
radius of the cylinder = 0.33 m
Tangential force = 210 N
angular acceleration = 0.978 rad/s²
we know,
[tex]\tau = F \times r[/tex]
and
[tex]\tau =I \times \alpha[/tex]
computing both the torque equation together
[tex]F \times r = I \times \alpha[/tex]
[tex]I = \dfrac{F\times r}{\alpha}[/tex]
[tex]I = \dfrac{210\times 0.33}{0.978}[/tex]
I = 70.86 kg.m²
Moment of inertia of the wheel is equal to 70.86 Kg.m²
