Question

A thin meter stick of mass 150 g rotates around an axis perpendicular to the stick’s long axis at an angular velocity of 240 rev/min. What is the angular momentum of the stick if the rotation axis (a) passes through the center of the stick? (b) Passes through one end of the stick?

a) (1.20 × 10⁻³ {kg} ⋅ {m}²/{s})
b) (1.44 × 10⁻³ {kg} ⋅ {m}²/{s})
c) (2.88 × 10⁻³ {kg} ⋅ {m}²/{s})
d) (4.32 × 10⁻³ {kg} ⋅ {m}²/{s})

Answer 1

To calculate the angular momentum of a rotating stick, we can use the formula: Angular momentum (L) = moment of inertia (I) * angular velocity (ω). The moment of inertia of the stick can be calculated using the formula for a thin rod. We can calculate the angular momentum for both scenarios where the rotation axis passes through the center of the stick and one end of the stick.

Step-by-step explanation:

To calculate the angular momentum of the meter stick, we can use the formula:

Angular momentum (L) = moment of inertia (I) * angular velocity (ω)

The moment of inertia of the meter stick can be calculated using the formula for a thin rod:

Moment of inertia (I) = (1/3) * mass * length^2

For the given problem, the mass of the stick is 150g and the length is 1m. Converting the mass to kg gives 0.15kg. Substituting these values into the formulas:

(a) If the rotation axis passes through the center of the stick:

Moment of inertia (I) = (1/3) * 0.15kg * (1m)^2

Angular momentum (L) = (1/3) * 0.15kg * (1m)^2 * (240 rev/min * 2π rad/rev * 1 min/60s)

(b) If the rotation axis passes through one end of the stick:

Moment of inertia (I) = (1/3) * 0.15kg * (1m)^2 + (0.15kg * (0.5m)^2)

Angular momentum (L) = (1/3) * 0.15kg * (1m)^2 * (240 rev/min * 2π rad/rev * 1 min/60s)

Calculating these values will give you the angular momentum of the stick in each scenario.