A sanding disk with rotational inertia 0.0012 kg-m2 is attached to an electric drill whose motor delivers a torque of magnitude 13 N-m about the central axis of the disk. About that axis and with a torque applied for 39 millisecond, what is the magnitude of the angular momentum?
To find the magnitude of the angular momentum, we need to calculate the angular velocity first. Angular velocity is the rate at which an object rotates around an axis and is given by the formula:
angular velocity (ω) = torque (τ) / rotational inertia (I)
In this case, the torque (τ) is given as 13 N-m and the rotational inertia (I) is given as 0.0012 kg-m². Plugging these values into the formula, we can calculate the angular velocity:
ω = 13 N-m / 0.0012 kg-m²
ω ≈ 10833.33 rad/s
Once we have the angular velocity, we can calculate the magnitude of the angular momentum using the formula:
angular momentum (L) = rotational inertia (I) * angular velocity (ω)
Plugging in the values, we get:
L = 0.0012 kg-m² * 10833.33 rad/s
L ≈ 13 N-m-s
Therefore, the magnitude of the angular momentum is approximately 13 N-m-s.