# Effect of load and load distance on peak and avg power

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.

JBA
Gold Member
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.

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.

JBA
Gold Member
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.

• Mark-01
There is one more thing that I want to ask: does the horizontal component of the motion need not be considered?

JBA
Gold Member
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.

CWatters
Homework Helper
Gold Member
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.

CWatters
Homework Helper
Gold Member
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.

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.

CWatters