Final answer:
The angular momentum of a 3.2-kg cylindrical grinding wheel with a 25 cm radius rotating at 1400 rpm is 14.66 kg·m²/s. The moment of inertia is 0.1 kg·m². Assuming no net torque is applied to change the angular velocity, the applied torque would be 0 N·m.
Step-by-step explanation:
To calculate the angular momentum (L) of a rotating cylindrical grinding wheel, you can use the formula L = I⋅ω, where I is the moment of inertia and ω is the angular velocity. For a uniform cylindrical object, the moment of inertia (I) is given by ½MR², where M is the mass and R is the radius. The angular velocity ω in radians per second can be found by converting the rotational speed from revolutions per minute (rpm) to radians per second using the conversion factor 2π radians/rev and 1 minute/60 seconds.
Given a mass (M) of 3.2 kg and radius (R) of 25 cm (0.25 m), and rotational speed of 1400 rpm, first calculate the moment of inertia:
I = 0.5 ⋅ 3.2 kg ⋅ (0.25 m)² = 0.1 kg·m².
Then, convert the angular speed to radians per second:
ω = 1400 rev/min ⋅ (2π radians/rev) ⋅ (1 min/60 s) ≈ 146.61 rad/s.
Finally, calculate the angular momentum:
L = 0.1 kg·m² ⋅ 146.61 rad/s = 14.66 kg·m²/s.
To address the torque, since it is not stated that any external torques are being applied to change the angular velocity, we assume the system is in steady-state rotation, which implies that there is no net torque applied to the system, hence torque would be 0 N·m.