1248.68 whats?Pinon1977 said:TL;DR Summary: Trying to determine how much power is created by a system that is rotating
So I have a system in which there is a disc with a moment of inertia of 1248.68.
Given the great specificity of the question, I'd say the specific answer is "some".Pinon1977 said:How would I go about determining how much power is exerted to do said work?
Let us go through the calculation. @russ_watters has already addressed the case of constant torque. Let us make no such assumption and compute average power.Filip Larsen said:Assuming your moment of inertia is in SI units you would need at least an average power of 4.4 kW to spin up the disc. But otherwise, yeah, what the other guys said!
Now that's what I call a GREAT answer and explanation. No sarcasm, no bull, just facts and figures. Thank you sir for this explanation.....it was rewarding and refreshing! And, most importantly, I learned something from it. Principal lesson learned is to get my units consistent. Haha.jbriggs444 said:Let us go through the calculation. @russ_watters has already addressed the case of constant torque. Let us make no such assumption and compute average power.
[I will not assume SI units, but instead simply assume a coherent system of units. But since we are given inputs involving minutes and seconds we will require our system of units to use one or the other]
Average power is computed simply. It is the change in energy divided by the length of the interval:$$P = \frac{\Delta E}{\Delta t}$$[In a coherent system of units, the unit of power will be chosen to match the unit of energy divided by the unit of time, so that no unit conversion constant appears in the formula above]
So we just have to figure out how much energy there is in an object with the given moment of inertia that is rotating at the given rate.
The rotational kinetic energy in an object with moment of inertia ##I## and rotation rate ##\omega## is given by $$E = \frac{1}{2}I\omega^2$$[In a coherent system of units, the unit of rotation rate will be a radian per time unit. The units for energy and for moment of inertia will work out so that no additional unit conversion constant appears in the above equation]
We need to convert 36 RPM to radians per second. We can proceed with the "multiply by one" method:$$36\text{ RPM} \times \frac{2\pi \text{ radians}}{1 \text{ rotation}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} = 3.77 \frac{\text{radians}}{\text{second}}$$[For the next step, we will need to assume that our system of units uses the second as its unit of time and, hence, radians per second as its unit of rotation rate]
We can now use the formula above (##E = \frac{1}{2}I\omega^2)## and plug in the moment of inertia of 1248.68 and the rotation rate of 3.77 radians per second to get:$$E = \frac{1}{2}(1248.68)(3.77)^2 = 8873$$[Of course, this result is in our system's chosen units of energy]
We are told that the acceleration takes place over the course of an elapsed time of 2 seconds. So:$$P = \frac{\Delta E}{\Delta t} = \frac{8873}{2} \approx 4400$$[If our system of units turns out to be SI then this is 4.4 kW, just as @Filip Larsen had calculated]
Instantaneous power in a rotating system with variable speed is calculated using the formula: \( P(t) = T(t) \cdot \omega(t) \), where \( P(t) \) is the instantaneous power, \( T(t) \) is the instantaneous torque, and \( \omega(t) \) is the instantaneous angular velocity.
Torque is typically measured in Newton-meters (Nm), and angular velocity is measured in radians per second (rad/s). When these units are used, the resulting power is in Watts (W).
Torque can be measured using a torque sensor or transducer, while angular velocity can be measured using a tachometer or an encoder. These devices provide real-time data that can be used to calculate instantaneous power.
Angular acceleration itself is not directly used to calculate power, but it affects the torque. The relationship is given by \( T(t) = I \cdot \alpha(t) \), where \( I \) is the moment of inertia and \( \alpha(t) \) is the angular acceleration. This torque can then be used in the power calculation.
To account for losses, you need to consider factors like friction, air resistance, and inefficiencies in the system. These losses can be quantified and subtracted from the calculated power to get the net power output. The formula becomes \( P_{net}(t) = T(t) \cdot \omega(t) - P_{loss}(t) \), where \( P_{loss}(t) \) represents the power losses at time \( t \).