1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rotary Energy Problem

  1. Jan 18, 2008 #1
    I am in the process of designing a renewable energy generator, but am having difficulty with some of the math’s, so if anyone could help me out, it would be appreciated. I have never studied advanced math’s or physics before, so please describe your explanation in layman’s terms. One of the problems I have come across is as follows:

    The Problem

    Look at Figure 1 (Attached)
    It is pivoted at C1 and C2
    It is 11m long and has a radius of 1 meter
    The weight of the empty semi-cylinder is 4 tonnes
    The semi-cylinder is full of fine sand
    The density of the fine sand is 2.0 tonnes / meter³
    The angle y is 20°
    How much power would it take to twist the semi-cylinder 20° in 3.5 seconds
    (Assume semi-cylinder is at sea level on planet earth)


    Relevant Equations

    Weight of Cylinder = pie.radius²(length.density)
    Force = Weight.Gravity
    Work = Force.Distance
    Power = Work / Time
    Acceleration = Distance / time²
    Force = mass.acceleration
    Torque = r(Force)
    Work = Angle.Torque
    Power = Work / time


    The Attempt

    Ok here is my calculation, but I don’t know if I’m right!
    I think there are three stages to this calculation
    Firstly: calculate the weight of the semi cylinder
    Secondly: calculate the potential energy
    Thirdly: calculate the inertia energy


    Semi Cylinder Filled with Fine Sand

    Density Fine Sand = 2.0 tonnes / meter³
    Radius = 1.00 meter
    Length = 11.00 meters

    Weight (Contents) = ½ π radius²(length x density)
    = 0.5 x 3.14 x 1 x 11 x 2
    = 34 tonnes (34000 kilograms)

    Weight (empty) = 4 tonnes (4000 kilograms)
    Weight (full) = 38t (38000kg)


    Potential Energy Calculation

    Weight displaced = Weight(y / 180)
    = 38000 x 20 /180
    = 4222 kg

    Force = Weight x Gravity
    = 4222 x 9.8
    = 41376 Newtons


    Distance of ‘Centre of Force’ from centre = r
    = 0.7 meters (assumed)

    Distance Centre of Gravity moves = pie.diameter (y / 360)r
    = 3.14 x 2 x (20 / 360) x 0.7
    = 0.244 meters

    Work = Force x Distance
    = 41376 x 0.244
    = 10096 joules

    Power = Work / Time
    = 10096 / 3.5
    = 2885 Watts
    = 2.885 Kilowatts


    Inertia Energy Calculation

    Distance Centre of Gravity moves = pie.diameter (y / 360)r
    = 3.14 x 2 x (20 / 360) x 0.7
    = 0.244 meters

    Acceleration = Distance / time²
    = 0.244 / 3.5²
    = 0.02 m/s²

    Force = mass x acceleration
    = 38000 x 0.02
    = 760 Newtons

    Dist of ‘Centre of Force’ from centre = r
    = 0.7 (assumed)

    Torque = r(Force)
    = 0.7 x 760
    = 532 NewtonMeters

     = y / 57.3
     = 20 / 57.3 radians
    = 0.349 radians

    Work = Angle(Torque)
    = 0.349 x 532
    = 186 joules

    Power = Work / time
    = 186 / 3.5
    = 53 watts
    = 0.053 kilowatts

    Total Power = Potential + Inertia
    = 2.885 + 0.053 (kW)
    = 2.938 kilowatts


    Any help appreciated
     

    Attached Files:

  2. jcsd
  3. Jan 19, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You're 'calculation' is missing any notion of trig functions and substituting made up numbers for them. Like in the "Weight displaced" line. I don't know you consider them advanced math, but you really need them for this problem. What really counts here is change in vertical elevation of the center of mass of the cylinder as the angle changes from one angle to another. Now, you don't have to 'assume' the location of the center of mass. It's the same calc as for a semicircle. You can look it up. The center of mass is 4/(3*pi) below the axis when the half cylinder is level. At a general angle y, it's cos(y)*4/(3*pi). So rotating it by 20 degrees takes different amounts of energy depending on the initial angle. If you want a good estimate of the maximum power rotate from say 80 to 100 degrees, since the maximum power is required at 90 degrees.
     
  4. Jan 20, 2008 #3
    Hey Dick

    Thanks for your help

    I’ll have a go at calculating it again taking into account what you said

    Actually I’m interested in all angles, but I need to know how to do this example first. I can then make a spreadsheet program and be able to swap around the variables

    Hope you get a chance to look at my second effort when I post it

    Inda
     
  5. Jan 22, 2008 #4
    OK HERES MY CALCULATION AGAIN:
    THIS TIME I THINK THERE ARE FOUR STAGES TO THIS CALCULATION
    FIRSTLY: CALCULATE THE WEIGHT OF THE SEMI CYLINDER
    SECONDLY:CALCULATE THE POTENTIAL ENERGY
    THIRDLY: CALCULATE THE TRANSVERSE INERTIA ENERGY
    FOURTHLY:CALCULATE THE ROTARY INERTIA ENERGY


    Semi Cylinder Filled with Fine Sand

    Density (Fine Sand) = 2.0 tonnes / meter³
    Radius = 1.00 meter
    Length = 11.00 meters

    Weight (Contents) = ½ π radius²(length x density)
    = 0.5 x 3.14 x 1 x 11 x 2
    = 34 tonnes (34000 kilograms)

    Weight (empty) = 4 tonnes (4000 kilograms)
    Weight (full) = 38tonnes (38000 kilograms)


    Potential Energy Calculation

    d = dist between ‘C of G’ and C of circle
    = 4 r / 3 π
    = 0.425 meters

    adjacent = d(cos y)
    = 0.425 (cos 20)
    = 0.399 meters

    h = dist C of G is raised vertically
    = d – adjacent
    = 0.425 – 0.399
    = 0.17 meters

    Force = Weight x Gravity
    = 38000 x 9.81
    = 372780 Newtons

    Work = Force x h
    = 372780 x 0.17
    = 63373 joules

    Power = Work / Time
    = 63373 / 3.5
    = 18107 Watts
    = 18.107 Kilowatts


    Transverse Inertia Energy Calculation

    d = dist between ‘C of G’ and C of circle
    = 4 r / 3 π
    = 0.425 meters

    (Chord – straight line)
    (G1 to G2) ² = b² + c² - 2bc (Cos A)
    (G1 to G2) ² = 0.18 + 0.18 – 0.36 * 0.94
    (G1 to G2) ² = 0.02
    G1 to G2 = 0.141 meters

    Acceleration = Distance / time²
    = 0.141 / 3.5²
    = 0.012 m/s²

    Force = mass x acceleration
    = 38000 x 0.012
    = 456 Newtons

    Work = Force x Distance
    = 456 * 0.147
    = 67.032 Joules

    Power = Work / Time
    67.032 / 3.5
    19.152 Watts
    0.0195 Kilowatts


    Rotary Inertia Energy Calculation

    (Dist along circumference)
    G1 to G2 = 20 / 360 * 2 π r
    = 0.148 meters

    Acceleration = Distance / time²
    = 0.148 / 3.5²
    = 0.012 m/s²

    Force = mass x acceleration
    = 38000 x 0.012
    = 456 Newtons

    Torque = d(Force)
    = 0.425 x 456
    = 193.8 NewtonMeters

     = y / 57.3
     = 20 / 57.3 radians
    = 0.349 radians

    Work = (Torque)
    = 0.349 x 193.8
    = 67.636 joules

    Power = Work / time
    = 67.636 / 3.5
    = 19 watts
    = 0.019 kilowatts

    Total Power = Potential + Transverse Inertia + Rotary Inertia
    = 18.107 + 0.019 + 0.019
    = 18.145 kilowatts


    The potential energy calculation seems to be ok now, but I can’t believe that the inertia energy calculations return such small figures. Am I doing it correctly?
     
  6. Jan 22, 2008 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's about right. The 'rotary calculation' is only approximately correct, since torque isn't constant. Torque is low when the angle y is near 0, it becomes large when y is near 90 degrees. That's why your answer is so low.
     
  7. Jan 22, 2008 #6
    Just a big thanks Dick

    Inda
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Rotary Energy Problem
  1. Energy problem (Replies: 7)

Loading...