Effect of load and load distance on peak and avg power

In summary: Radius (r) = (Es-Ef)/(E-Ef) = (0.5 Msrs2-0.5 Ifs2)/(0.5 Msrs2+0.5 Ifs2) = (0.5 Msrs2+0.5 Ifs2)/(0.8 Msrs2+0.2 Ifs2)At this point the power is maximum.
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I am trying to create a lever which has mass on one end and i need to lift the lever. A picture showing the lever setup is in this link: https://imgur.com/WeH4oK7

The system is like seesaw ( from -20 degrees to +20 degrees at constant angular velocity for time period 't') The input angular velocity and effort end length is same while load and load distance is varied (though load * load distance is constant). I am confused by which method should I solve the problem to find required input power.

I have come up with this much of thinking:

1> Assuming radial acceleration is effective:

Power(P)
= Torque * angular velocity(omega)
= Force * force distance(r) * omega
= [m * (v^2 /r) ]* r * v/r
= m * v^3 /r

2> Neglecting radial acceleration since the horizontal component is small and the motion is not continuous circular but more like seesaw:

P = Work done / time taken
= F * d / t
= m * v/t * d/t
= m * v^2 /t

Which one should I use or if my approach has mistakes, kindly help me get through it.
 
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  • #2
First of all, there is no way to have a constant velocity in reciprocating lever i.e. the lever travels to upmost position, stops, reverses, travels to it bottom most position, stops and repeats.

Second, the first requirement is to determine the amount of force required to lift your load based upon the lever ratio and the vertical travel of the power end of the lever by simple geometry. Once you have those basics, the amount of power required = force x distance / time of travel.
 
  • #3
JBA said:
First of all, there is no way to have a constant velocity in reciprocating lever i.e. the lever travels to upmost position, stops, reverses, travels to it bottom most position, stops and repeats.

Second, the first requirement is to determine the amount of force required to lift your load based upon the lever ratio and the vertical travel of the power end of the lever by simple geometry. Once you have those basics, the amount of power required = force x distance / time of travel.
It is constant velocity at normal time except top and bottom position change.
Power is f*d/t but I am talking about instant power at the time between when body is at rest and the body starts to move up.
 
  • #4
In that situation the only thing that changes is the effect of the systems inertial resistance has to be added to the above load and for this you have to decide how quickly you want to reach your constant speed, i.e. the required acceleration rate. From that it is simply a matter F = mass of system x acceleration.
 
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  • #5
Thank you for your help.
There is one more thing that I want to ask: does the horizontal component of the motion need not be considered?
 
  • #6
As for any affects from horizontal forces that is a matter of how rigorous you want your calculation to be. Technically, all forces should be considered regardless of their potential level of contribution to the total loading on the system during operation.
 
  • #7
Mark-01 said:
The system is like seesaw ( from -20 degrees to +20 degrees at constant angular velocity for time period 't') The input angular velocity and effort end length is same while load and load distance is varied (though load * load distance is constant). I am confused by which method should I solve the problem to find required input power.

Here is my take on the problem...

The load torque is roughly constant at 10 * 9.81 * 1 = 98.1Nm and the angular velocity is constant. So the power required to overcome gravity is roughly...

Power (W) = Torque (Nm) * Angular Velocity (Rad/s)………………………..(1)

but this isn't the whole story. It ignores the fact that the moment of inertia (I) is changing. The moment of inertia of a point mass at radius r is

I = mr2

In this case m and r vary from start (s) to finish (f).

Is = msrs2
If = mfrf2

The energy stored in a rotating object is given by..

E = 0.5 I ω2

The energy at the start and finish is therefore also changing...

Es = 0.5 Isω2 = 0.5 msrs2ω2
Ef = 0.5 Ifω2 = 0.5 mfrf2ω2

The difference between these two values divided by time gives you the average power required by the changing angular momentum. Depending on the direction it could be negative. You have to add or subtract this from eqn (1) above to get the total power.

To calculate the peak power you have to figure out at what point the energy is changing the most rapidly. That depends on how the mass and radius are changing. Does the radius change at a constant rate? Does the mass change at a constant rate? They can't both change at a constant rate.
 
  • #8
If the radius changes linearly with the angle I believe the moment of inertia changes at a constant rate as well. So the peak power is the same as the average power.
 
  • #9
I'm confused. Why are you interested in power at all? In any oscillating system as pictured, every time the lever stops the power is zero. The power then increases as the rate of angular rotation increases, then goes to zero again at the end of the lever's stroke. The magnitude of the force dictates the angular acceleration of the lever. The force must be greater than the (load X load distance) to get the load moving "up". Then the force would have to decrease to allow the load to stop moving up. Then the force would have to be less than the (load X load distance) to allow the load to return to its "down" position.The magnitude of the driving force compared to the magnitude of the load would dictate the angular acceleration in both up and down movements, greater to "lift", less to "lower". So both force and power are constantly changing. You could calculate the force and power for a complete cycle, I suppose, but averaging that out will not result in useful information.
 
  • #10
The OP refers to constant angular velocity.
 
  • #11
Let's assume that the load is moving at some constant angular velocity through a single stroke. In that case there is no acceleration so the effort is constant and (effort X distance from effort to fulcrum) = (load X distance from load to fulcrum). So the power would be the (effort X distance traveled by the effort / time for one stroke). If the effort is continuous (implied by no acceleration) then the power would be continuous, a single value. That value would be the same as (load X distance traveled by the load / time for one stroke). Force X distance / time.
The radial component needn't be considered because of the geometry of levers.
 

1. What is the effect of load on peak power?

The amount of load placed on a system can impact its peak power output. As the load increases, the peak power output decreases due to the increased resistance and energy required to move the load. This is known as the load effect.

2. How does load distance affect peak power?

The distance at which the load is placed from the power source can also impact the peak power output. The farther the load is from the power source, the more energy is required to move it, resulting in a decrease in peak power output.

3. What is the relationship between load and average power?

The relationship between load and average power is inversely proportional. As the load increases, the average power decreases. This is due to the increased resistance and energy required to move the load, resulting in a decrease in the amount of power that can be sustained over time.

4. How does load distance affect average power?

Similar to peak power, the distance at which the load is placed from the power source can also impact the average power output. As the load distance increases, the average power output decreases due to the increased energy required to sustain the load over a longer distance.

5. What is the significance of understanding the effect of load and load distance on peak and average power?

Understanding the effect of load and load distance on peak and average power is important for optimizing the performance of a system. By knowing how the load and load distance impact power output, scientists and engineers can design more efficient systems and make informed decisions about load placement to achieve desired power outputs.

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