A particle of mass 20 gm is moving in a circle of 4cm radius with contant speed of 10 cm/sec. What is the angular momentum (gm. cm2 / sec) about a point on the axis of the circle 3 cm distance from the centre.Correct answer is '1000'. Can you explain this answer?

Question

Answer

Given:
Mass of the particle, m = 20 gm = 0.02 kg
Radius of the circle, r = 4 cm = 0.04 m
Speed of the particle, v = 10 cm/sec = 0.1 m/sec
Distance of the point on the axis from the center, d = 3 cm = 0.03 m

To find:
Angular momentum of the particle about the given point.

Solution:

1. Angular Velocity:
The particle is moving in a circle with a constant speed, which means it has a constant tangential velocity. The tangential velocity is given by v = rω, where ω is the angular velocity.

Substituting the given values, we have:
0.1 m/sec = 0.04 m × ω

Simplifying the equation, we get:
ω = 0.1 m/sec / 0.04 m = 2.5 rad/sec

2. Angular Momentum:
The angular momentum of a particle about a point is given by L = Iω, where I is the moment of inertia and ω is the angular velocity.

The moment of inertia of a particle moving in a circle about an axis is given by I = mr^2.

Substituting the given values, we have:
I = 0.02 kg × (0.04 m)^2 = 0.02 kg × 0.0016 m^2 = 0.000032 kg·m^2

Substituting the values of I and ω, we have:
L = 0.000032 kg·m^2 × 2.5 rad/sec = 0.00008 kg·m^2·rad/sec

3. Converting to gm.cm^2/sec:
To convert the angular momentum to gm·cm^2/sec, we need to multiply by 1000.

L (in gm·cm^2/sec) = 0.00008 kg·m^2·rad/sec × 1000 gm/kg × 100 cm/m = 8 gm·cm^2/sec

4. Angular Momentum about the given point:
To calculate the angular momentum about the given point, we need to consider the perpendicular distance between the point and the axis of rotation.

The perpendicular distance, r', is given by r' = r - d = 0.04 m - 0.03 m = 0.01 m

Substituting the values of r' and ω, we have:
L = Iω = (mr'^2)ω = (0.02 kg × (0.01 m)^2) × 2.5 rad/sec = 0.02 kg × 0.0001 m^2 × 2.5 rad/sec = 0.00005 kg·m^2·rad/sec

Converting to gm·cm^2/sec, we have:
L (in gm·cm^2/sec) = 0.00005 kg·m^2·rad/sec × 1000 gm/kg × 100 cm/m = 5 gm·cm^2/sec

Therefore, the angular momentum of the particle about the given point is 5 gm·cm^2/sec.

Answer

Angular momentum about a point on the axis of the circle

Since,

is perpendocular to the velocity

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