A thin ring has mass 0.25 kg and radius 0.5 m its moment of inertia about an axis passing through its centre and perpendicular to its plane is
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moment of inertia of ring about axis passing through its centre and perpendicular to its plane is given by
I = mr²
where m is the mass of ring and r is the minimum seperation between axis of rotation and point of observation { e.g., radius of ring }
here, m = 0.25 Kg
r = 0.5 m
now, I = 0.25 × (0.5)² = 0.25 × 0.25 = 0.0625 Kg.m²
I = mr²
where m is the mass of ring and r is the minimum seperation between axis of rotation and point of observation { e.g., radius of ring }
here, m = 0.25 Kg
r = 0.5 m
now, I = 0.25 × (0.5)² = 0.25 × 0.25 = 0.0625 Kg.m²
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