Consider a circular ring of mass 1 kg and diameter 0.2 m. It is making 10 rotations per second about an axis passing through its centre and normal to the surface. The value of angular momentum is:-

Concept:

• Angular momentum (L) or rotational momentum is the moment of inertia times (I) angular velocity (ω).
• Angular momentum is the Vector quantity and its directly proportional to the toque applied to the body.
• When the direction of torque and direction of rotation is the same then angular momentum increases and when the direction of torque and rotation is opposite then angular momentum decreases.
   

∴ L = I × ω

Where,

L is angular momentum
and, ω is angular velocity

Explanation:

Given data,

Circular ring of mass (m) = 1 kg,

diameter (d) = 0.2 m → radius (r) = 0.1 m

• As we know the angular momentum of a ring for which axes of rotation pass from its center and normal to the plane is I = mr²
• As the ring rotates 10 revolutions per second it means it has angular velocity (ω) = 10 × 2π \(\frac{rad}{s}\)
   

Now Angular momentum (L) = I × ω 

∴ L = mr² × ω

∴ L = (1) ⋅ (0.1)² × 20π 

∴ L = 0.628 kg m² s^-1

Hence the angular momentum of the circular ring becomes 0.628 kg m² s^-1.

Note: This question was deleted in RSMSSB Lab Assistant 2016 Exam