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What is the angular momentum of a 3.2- kg uniform cylindrical grinding wheel of radius 25 cm when rotating at 1300 rpm ?

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Final answer:

The angular momentum of a 3.2-kg cylindrical grinding wheel of radius 25 cm rotating at 1300 rpm is approximately 13.6 kg⋅m²/s, calculated using the moment of inertia for a solid cylinder and converting the angular velocity to radians per second.

Step-by-step explanation:

To find the angular momentum of a 3.2-kg uniform cylindrical grinding wheel with a radius of 25 cm rotating at 1300 rpm, we can use the formula for the angular momentum (L) of a rotating body:

L = I⋅ω

where:

  • I is the moment of inertia of the cylinder, given by ½MR² for a solid cylinder.

First, we calculate the angular velocity in rad/s:

ω = 1300 revolutions/min × (2π rad/revolution) / (60 s/min) ≈ 136.1 rad/s

I = ½ × 3.2 kg × (0.25 m)² ≈ 0.1 kg⋅m²

Now we can compute the angular momentum:

L = 0.1 kg⋅m² × 136.1 rad/s ≈ 13.6 kg⋅m²/s

So, the angular momentum of the grinding wheel is approximately 13.6 kg⋅m²/s.