• Welcome to engineeringclicks.com
  • Calculating HP to spin wheel

    Discussion in 'Calculations' started by davidhiep86, Dec 22, 2014.

    1. davidhiep86

      davidhiep86 Member

      Dec 2014
      Likes Received:
      I've been tasked with calculating the HP required to spin a turntable from rest to a desired RPM. I've tried calculating this out using two different methods, and have came upon two different solutions. I've went through my formulas multiple times but have not found why I am getting two different answers (one is double of the other).

      r = radius of turntable [m]
      m = mass of turntable [kg]
      ω = desired angular speed [RPM]; converted to rad/s using 2*pi/60.
      t = desired time

      Method 1: Energy

      Erot = 1/2 * I * ω^2

      This gives me my energy required in Joules which I then divide by the time to get my required power. Using the moment of inertia of a think disc (1/2*m*r^2), my final equation looks like this:

      P = [(1/2)*(1/2)(mr^2)*ω^2]/t
      = [(1/4)*(mr^2)*ω^2]/t

      Method 2

      For the second method, I used the equations below:

      τ = I * α
      P = τ * ω ==> P = I * α * ω

      First I calculated my angular acceleration.

      α = (ωf - ωi)/t = ωf/t

      The equation simplifies to ωf/t since we are starting from rest.

      I then input this into my power equation using the moment of inertia of a disc once again. This time I get:

      P = (1/2)*(mr^2)*(ωf/t)*ωf
      = [(1/2)*(mr^2)*ωf^2]/t

      As you can see, my two results differ by the number in the denominator. I have the first equation being multiplied by (1/4) and the second multipled by (1/2). Does anyone know if I am missing any steps in one of these equations that might cause me to miss there being another 2 being multiplied. Any help would be appreciated!
    3. Aravind V

      Aravind V New Member

      Jul 2013
      Likes Received:
      First, angular velocity. omega in radians per second= take (2*pi*RPM(rev/min)/60

      As for HP, first of all you need calculate the moment of inertia for the disk i.e. I=.5mr^2. Notice r=radius
      Then, the rotational energy = .5*I*omega^2.

      That should give you the energy required to give that disc that speed.
      To find the required power, divide that energy by the time given and you will have Joules/sec, the unit of power equal to a Watt. Then just convert to HP.
    4. mike_atlas

      mike_atlas New Member

      Aug 2013
      Likes Received:
      I believe you are confusing Average power with instantaneous power.
      The Average power (total energy divided by time) is as you calculated.

      The power required at any particular time is also as you calculated: P = I * α * ω

      However, ω is not a constant. For example, when you first start the wheel going, the power needed is basically 0 as the speed ω is 0.

      If you plotted power vs. speed it would be a straight line starting at 0 and ending at P = I * α * ω . The integral of this (area under the line) would be 1/2 the final power, or the same as calculated with the energy equation.
    5. Reddog1

      Reddog1 New Member

      Jan 2015
      Likes Received:
      I see that this thread is still open. I'd like to make a couple of comments that might help your analysis. They are in regards to your basic view of the problem. I can see where your equations differ, but they are both wrong in principle. Some of your responses have picked up on part of the basic flaws but are not fully explored. I will attempt to explain in more detail. Please don't get offended if I start with basics.

      You are really dealing with torque during the startup. While Hp is an average energy(work) over time. And the energy analysis gives an indication of the answer. It does not give an exact answer.

      Consider that a 1 hp motor that operates at 100 rpm has a relatively high torque value. While a 1 hp motor that operates at 100,000 rpm has much lower torque value. The motors can be geared up or down. (With losses of course) But torque is the concept to consider.

      Is your motor direct drive, belt or gears? This is one consideration.

      What would your torque curve for the motor look like?

      What are your friction losses at speed?

      Is power consumption or rpm stability significant?

      DC or AC or electronic stepper motor?

      All of these things can affect your answer to find the right motor.

      An ideal answer must consider if the spinup is linear or not. This depends on the torque of the motor system throughout the runup.

      What you have been calculating is the work energy expended during the spinup time. This does not really tell you what motor is required. It just gives you clues about the system.

      Ignoring friction and system losses for the moment. Start with the basic F=MA law applied to rotation.

      Assume constant angular acceleration for the moment. Then consider that your final angular velocity, start angular velocity and required time to get a value for the required angular acceleration. This will allow you to compute the required torque from the basic law. (naturally, you are using moment of inertia for the computation- not just mass) (you already have done most of this analysis)

      Look for a motor that will give you the necessary torque in your overall system without overkill. At speed, there will probably not be a lot of friction loss. The motor hp will come from the necessary torque value, not from a straight power conversion formula.

      Now go back and consider that the motor you are looking at, does not produce a constant torque as rpm changes. Message your analysis to more closely fit the actual motor system (and your motor accordingly). Keep in mind that you must also consider the moment of inertia of the motor(belts,gears,bearings,etc.).

      A power conversion formula for motor hp works for steady state situations(with everything considered), but is rarely accurate for rate change situations. I also think it is worth noting, that motor hp is given at speed, not at a partial motor speed. Unless there are extenuating circumstances. You must consider the torque curve of the motor system.

    Share This Page