Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Torque and power

  1. Nov 26, 2014 #1
    Hallo
    Currently I am engaged with my Company's Project and it is about Electric motor as my Knowledge is not that good about Electric motors, I need your help. I want to move 80 kgs of Weight with the speed of 80 km/h with a motor and a 260 mm pulley. The motor is standing vertically and the pulley above the floor. Now I need to calculate how much Torque and Power I need for my Motor so that I can move the weight (80 Kgs) with 80 km/h speed. Any help would be really appreciated.

    Thanking You
    Prem
     
  2. jcsd
  3. Nov 26, 2014 #2
    To calculate the torque required its simply 1/2*D*L where D is the pulley diameter and L is the load.
     
  4. Nov 26, 2014 #3
    Ya but I want to calculate Torque and Power to move 80 kgs of weight with 80 km/h
     
  5. Nov 26, 2014 #4
    I would double check this but in English units HP= torque * rpm ÷5252. So if you calculate the required torque, you know the HP of the motor you find rpm. So translate you required speed to rpm and I think you got it.
     
  6. Nov 26, 2014 #5
    Since velocity is a linear measure (distance of time) and you are spinning a shaft which is angular velocity you have to relate them by converting velocity to RPM. Hope this helps!
     
  7. Nov 26, 2014 #6
    Ya I understood the formula but the thing is rpm is revolution per minute (rotational motion) and velocity in linear velocity and I am not sure if calculating rpm will directly give me linear velocity that I need to move the object
     
  8. Nov 26, 2014 #7
    you have to convert velocity to angular velocity to when you are using a spinning shaft. Angular velocity is calculated by v = r × ω
    Where:
    v: Linear velocity, in m/s
    r: Radius, in meter
    ω: Angular velocity, in rad/s

    To do this in RPM you would use v = r × RPM × 0.10472

    Where:
    v: Linear velocity, in m/s
    r: Radius, in meter
    RPM: Angular velocity, in RPM (Rotations per Minute)
     
  9. Nov 27, 2014 #8

    billy_joule

    User Avatar
    Science Advisor

    The answer lies in two factors:
    How fast do you want to get to 80km/hr
    What forces are acting on the mass at steady state - There will likely be significant air drag at 80km/hr among other sources of friction.

    The mass of the cable makes things a little difficult - the moment of inertia of the pulley system increases with time as the mass of the cable is probably not negligible.
     
  10. Mar 26, 2016 #9
    Your question is incomplete. For example, are you lifting the weight or moving it horizontally? Let's assume you are lifting the weight. The torque required for a constant lifting speed would be (pulley radius X (load weight + system friction). The starting torque depends on how quickly you want to reach the required 80 km/h speed, that is, how quickly must you accelerate the load? That would be {[(axle torque / pulley radius) - (load weight + starting friction] / mass of the load}.

    This is pretty sparse. Post more if this doesn't help.
     
  11. Mar 29, 2016 #10

    David Lewis

    User Avatar
    Gold Member

    The relationship between torque and power is power = speed x torque.
    This holds true no matter what units you use, but if you employ an inconsistent system of units, you'll need to insert conversion factors.
    The following units are standard:
    [power] = watts (W)
    [speed] = radians per second (rad/s)
    [torque] = newton-metres (N-m)
     
  12. Mar 30, 2016 #11
    Just an editorial comment, I do wish "power" wasn't used so much in mechanical engineering. It is misunderstood by so many people, including engineers. Power isn't a "thing", it's a rate. It doesn't do anything. For example, power cannot be used to calculate the rate of acceleration on a mass or how sturdy a driveshaft must be or the loading on a gear tooth, etc. For any given power and mass there are an infinite number of valid acceleration values. For any rate of power delivered by a driveshaft there are an infinite number of valid torque values. Conversely, for any given mass and rate of constant acceleration there are an infinite number of valid power values. Etc., etc.
     
  13. Apr 1, 2016 #12
    Power, hp = Workdone in kg-m per S / 75 = 80 kg * (80,000/3600 m/s) / 75 = 23.70 hp

    Also,
    Power = (2*Pi*N*T) / 75
    Torque, T = (Power *75) / (2 * Pi * N)

    N (rps) = Linear Velocity / (Pi * Diameter) = (80,000/3600 m/s)/ (Pi * 0.26 m) = 27.19 rps

    Torque, T = (23.70 hp * 75) / (2 *3.143 *27.19 rps) = 10.40 kg-m

    A N Madhavan
     
  14. Apr 1, 2016 #13

    billy_joule

    User Avatar
    Science Advisor

    Do we need to keep adding to this thread? The OP hasn't replied since over four months ago.
     
  15. Apr 1, 2016 #14
    I just wanted to brush-up my basics, so responded and probably it would help someone.
     
  16. Apr 1, 2016 #15

    billy_joule

    User Avatar
    Science Advisor

    This is not right.
    Power = force*velocity
    We don't know the force required to maintain the constant 80km/hr, as mentioned in post #8 so we cannot calculate power.
     
  17. Apr 1, 2016 #16
    B
    Both are same.
    Force (kg) * Velocity (m/s) = Work done per S in kg-m/s
    75 kg-m/s = 1 hp
    The work done to move 80 kg at (80,000/3600 m/S) is 80 * 22.22 kg-m/s => 1777.77 kg-m/s => 23.70 hp
     
  18. Apr 1, 2016 #17

    billy_joule

    User Avatar
    Science Advisor

    Force is not measured in kg.
    We don't know the force, the object is moving horizontally (post #1 implied this).
    Your approach would apply if the weight was being lifted vertically.
     
  19. Apr 1, 2016 #18
    That is true.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Torque and power
  1. Torque and power. (Replies: 4)

Loading...