To tackle this problem, we need to analyze the dynamics of the ultracentrifuge rotor, focusing on its angular motion. We'll break this down step by step, calculating the necessary braking torque and the frictional torque present during normal operation.
Understanding the Rotor's Dynamics
The rotor can be modeled as a uniform cylinder, which allows us to use the equations of rotational motion. The key parameters we have are:
- Radius (r) = 10 cm = 0.1 m
- Mass (m) = 8 kg
- Initial speed (ω_initial) = 80,000 rpm
- Time to stop (t_stop) = 30 s
- Time to stop without braking (t_no_brake) = 30 min = 1800 s
Converting RPM to Radians per Second
First, we need to convert the initial speed from revolutions per minute (rpm) to radians per second (rad/s). The conversion factor is:
1 revolution = 2π radians, and 1 minute = 60 seconds.
Thus, the conversion is:
ω_initial = 80,000 rpm × (2π rad / 1 rev) × (1 min / 60 s) = 8,377.58 rad/s
Calculating Moment of Inertia
The moment of inertia (I) for a uniform cylinder rotating about its symmetry axis is given by:
I = (1/2) m r²
Substituting the values:
I = (1/2) × 8 kg × (0.1 m)² = 0.04 kg·m²
Finding Angular Deceleration
Next, we need to determine the angular deceleration (α) required to stop the rotor in 30 seconds. We can use the formula:
ω_final = ω_initial + α t
Since the final angular velocity (ω_final) is 0 (the rotor stops), we rearrange the equation:
0 = 8,377.58 rad/s + α (30 s)
Solving for α gives:
α = -8,377.58 rad/s / 30 s = -279.25 rad/s²
Calculating the Minimum Braking Torque
The torque (τ) required to produce this angular deceleration can be calculated using the formula:
τ = I α
Substituting the values we have:
τ = 0.04 kg·m² × (-279.25 rad/s²) = -11.17 N·m
The negative sign indicates that this torque is applied in the opposite direction of the rotor's motion. Thus, the minimum braking torque required is approximately 11.17 N·m.
Determining the Frictional Torque
To find the frictional torque present under normal spinning conditions, we can use the information about the time it takes to stop without any braking torque. If the rotor stops in 30 minutes (1800 seconds) due to friction, we can find the frictional torque using a similar approach.
Using the same angular deceleration formula:
0 = 8,377.58 rad/s + α_friction (1800 s)
Solving for α_friction gives:
α_friction = -8,377.58 rad/s / 1800 s = -4.65 rad/s²
Now, we can calculate the frictional torque:
τ_friction = I α_friction
Substituting the values:
τ_friction = 0.04 kg·m² × (-4.65 rad/s²) = -0.186 N·m
This indicates that the frictional torque acting on the rotor during normal operation is approximately 0.186 N·m.
Summary of Findings
In summary, the minimum braking torque required to stop the rotor in under 30 seconds is about 11.17 N·m, while the frictional torque present during normal spinning conditions is approximately 0.186 N·m. Understanding these values is crucial for the effective operation and safety of ultracentrifuge systems.
