Final answer:
The angular momentum of a 4.0-kg uniform cylindrical grinding wheel with a radius of 23 cm rotating at 1300 rpm is approximately 72.006 kg·m²/s.
Step-by-step explanation:
To calculate the angular momentum (L) of a uniform cylindrical grinding wheel, we can use the following relationship:
L = I × ω
where I is the moment of inertia of the cylinder, and ω (omega) is the angular velocity in radians per second.
For a uniform solid cylinder, moment of inertia (I) is given by:
I = ½MR²
where M is the mass and R is the radius.
The angular velocity (ω) in radians per second can be found by converting rpm (rotations per minute) into radians per second (1 rpm = 2π/60 radians per second).
ω = 1300 rpm × (2π rad/1 rev) × (1 min/60 s)
Now, substituting the values:
M = 4.0 kg, R = 0.23 m and ω in radians per second.
First, compute ω:
ω = 1300 × (2π/60) ≈ 136.13 rad/s
Then calculate I:
I = ½ × 4.0 kg × (0.23 m)² ≈ 0.529 kg·m²
Finally, L can be calculated:
L = 0.529 kg·m² × 136.13 rad/s ≈ 72.006 kg·m²/s
Therefore, the angular momentum of the grinding wheel is approximately 72.006 kg·m·s-1.