# Rotary Energy Problem

1. Jan 18, 2008

### Inda House

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

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
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

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:

• ###### Fig 1- Semi Cylinder.jpg
File size:
17.1 KB
Views:
69
2. Jan 19, 2008

### Dick

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.

3. Jan 20, 2008

### Inda House

Hey Dick

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

4. Jan 22, 2008

### Inda House

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
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

= 0.425 (cos 20)
= 0.399 meters

h = dist C of G is raised vertically
= 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

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?

5. Jan 22, 2008

### Dick

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.

6. Jan 22, 2008

### Inda House

Just a big thanks Dick

Inda