A 0.2 kg particle is attached at the rim of rotating wheel of radius 0.5m. The wheel completes each revolution in 0.8s. The moment of inertia of the wheel (not including the particle is 0.16kg.m2. The magnitude of the angular momentum of the system is:a.1.65kg.m2/sb.3.65kg.m2/sc.165kg.m2/sd.2.65kg.m2/s
Question
A 0.2 kg particle is attached at the rim of rotating wheel of radius 0.5m. The wheel completes each revolution in 0.8s. The moment of inertia of the wheel (not including the particle is 0.16kg.m2. The magnitude of the angular momentum of the system is:a.1.65kg.m2/sb.3.65kg.m2/sc.165kg.m2/sd.2.65kg.m2/s
Solution
The angular momentum (L) of a system is given by the formula:
L = Iω
where: I = moment of inertia ω = angular velocity
The moment of inertia of the system is the sum of the moment of inertia of the wheel and the moment of inertia of the particle. The moment of inertia of the particle can be calculated using the formula:
I = mr²
where: m = mass of the particle r = radius of the wheel
Plugging in the given values:
I = 0.2 kg * (0.5 m)² I = 0.2 kg * 0.25 m² I = 0.05 kg.m²
The total moment of inertia of the system is:
I_total = I_wheel + I_particle I_total = 0.16 kg.m² + 0.05 kg.m² I_total = 0.21 kg.m²
The angular velocity can be calculated using the formula:
ω = 2π / T
where: T = period of rotation
Plugging in the given value:
ω = 2π / 0.8 s ω = 7.85 rad/s
Finally, we can calculate the angular momentum:
L = Iω L = 0.21 kg.m² * 7.85 rad/s L = 1.65 kg.m²/s
Therefore, the magnitude of the angular momentum of the system is 1.65 kg.m²/s.
The correct answer is:
a. 1.65kg.m²/s
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