Help Me With A Physics Concept

[jyl]

Help me understand a physics concept. I don't get the intuition behind it.

It has to do with power.

I'll ask my question in the context of a pulley problem.

A block of mass M1 is on a frictionless plane, inclined at angle Z. A string is fastened to that block. The string passes over a frictionless pulley. The other end of the string is fastened to a hanging block of mass M2.

The force of gravity on block 2 is F2 = M2 x g. The component of the force of gravity on block 1 that is tangent to the plane is F1 = M1 x g x sin Z.

If M2 > M1 x sin Z, then block 2 will descend and pull block 1 up the inclined plane. The blocks will accelerate at A = [ (M2 - M1 * sin Z) / (M1 + M2) ] * g although mathematically block 1 has acceleration A and block 2 has acceleration -A. The blocks will have the same scalar component of velocity V.

Assuming the blocks start at rest, at time T, velocity V = A * T.

(Please correct me if I have it wrong so far. This is straight out of first year physics but college was a long time ago.)

I'm trying to calculate the power that block 2 applies to block 1. Block 2 is applying a constant force F2 = M2 * g to block 1. Power = force x velocity. So Power P = F2 x V.

(Still right?)

Now, this is what I don't intuitively get. As T increases, with constant A, velocity increases. That means power P increases. At T = 0. P = 0. The faster the masses move, the higher the power that is being applied to block 1.

If the force remains constant, how can power increase?

I don't understand this intuitively.

Actually, I always have trouble intuitively understanding power. I recall vaguely that in physics class, we always calculated Work, and then divided by time to get Power. We didn't start by calculating Power. Maybe that's why I never really got comfortable with Power.

Using that approach, I can see that block 2's kinetic energy increases at a doubly accelerating rate, so that in the 1 second interval from T = 1 to 2 there is a small increase in KE, while in the 1 second interval from T = 10 to 11 there is a much larger increase in KE. So if P = KE / T, the power in the first interval is small while the power in the second interval is much larger. But I don't "get it".

[DanielDudley]

Gravity is a constant force. I am six feet tall, and I fall down and hit my head. Oh, and I am standing on a 100 foot tower. How could that hurt more than just falling down at ground level ?

[pavulon]

It won't hurt more.

Originally Posted by DanielDudley

Gravity is a constant force. I am six feet tall, and I fall down and hit my head. Oh, and I am standing on a 100 foot tower. How could that hurt more than just falling down at ground level ?

[onewhippedpuppy]

Force = mass * acceleration. In that equation, gravity is acceleration. That's where you get your increasing velocity.

Power = Work / Time = (Force * Displacement) / Time = Force * Velocity

You are combining a lot of different physics problems into one. What are you actually looking to calculate, and why? Power is frequently misunderstood, from the layman's perspective the use of power is often incorrectly substituted for torque to describe a force. Power is simply the rate of doing work, which is why it takes several calculations to get there.

[Arizona_928]

Originally Posted by onewhippedpuppy

Force = mass * acceleration. In that equation, gravity is acceleration. That's where you get your increasing velocity.

Power = Work / Time = (Force * Displacement) / Time = Force * Velocity

^^

I find it easier to follow the units (and gasp put in values) if I get confused half way through trying to derive the equation.
I'm not that good with newtonian physics though.

[Flieger]

Dyno chart for an engine: if torque is flat power increases linearly with speed. Simple.

In your thought experiment I'm not sure what you are trying to get but I think there is probably something slightly wrong with how you are looking at power in this situation because the gravitational potential energy is being converted into kinetic energy distributed in a certain way between the two components of the system.

[javadog]

Originally Posted by jyl

If the force remains constant, how can power increase?

Don't equate force, or work done, with power.

Power introduces an element of time.

That's where velocity enters the scene.

JR

[porsche4life]

My brain hurts... Thanks for reminding me why I stayed away from engineering!

[jyl]

Thanks. Here is the problem I am trying to solve. It isn't a practical project, just a mental hobby thing. There is a concept, I am trying to figure out if the physics would work.

The concept (which is not my idea, I don't take credit) is a wheel that contains an internal hollow channel, that is annular, circling the circumference of the wheel. Inside the channel, there is a weight. The weight can slide frictionlessly around the channel. There is some unspecified "mechanism" that allows the weight to be driven around the channel by applying force to the wheel (imagine a gear, or a solenoid). Set the vehicle on the ground, facing to your right, with the weight at 3 o'clock. The gravity pushes the weight down (clockwise) in the channel. Use the drive mechanism to hold the weight at 3 o'clock. The reaction force on the wheel will rotate the wheel clockwise and propel the vehicle forward. Relative to the wheel, the weight will be rotating counterclockwise. Relative to the vehicle, the weight will be stationary.

The vehicle weighs 85 kg (bicycle + rider). The weight weighs 2 kg. Radius of the wheel is 35 cm.

My initial reaction was that this drive system will produce very little force, just gravity force on a 2 kg mass, call it 20 N. I considered it essentially like a mass (the weight) hanging from a pulley, connected by the string to another mass (the vehicle). Hence the set-up I described in the original post. I calculated the force would only be able to propel the vehicle up a 1.3 degree slope.

The person who proposed the concept pointed out that the drive system would work as a primary drive on flat road, and as an auxiliary drive on inclines. On flat road, even 20 N will accelerate the vehicle from a dead stop. It is only a small acceleration, but since it is sustained, the vehicle will get to reasonable speeds before too long. On an incline, you would use another force (pedaling) as your primary power source to get the vehicle going, and the drive system provides additional force. He also said that the power provided by the drive system will increase as the vehicle accelerates. The claim is that by the time the vehicle is going a reasonable speed (10 m/s) the drive system is providing 200 watts of power (20 N x 10 m/s) which is actually a lot, relative to the primary power source (an average cyclist can only pedal at a 200 watts for an hour).

I am having trouble grasping this intuitively. I can't grasp how a constant force of 20 N can produce ever-increasing power with increasing velocity, if the acceleration is due or partly due to another force.

The other thing is, I am starting to doubt my initial assumption that holding the weight at 3 o'clock produces a constant force of 20 N on the rim. With the wheel at rest, it does. But when the wheel is rotating fast?

Again, please ignore the practical aspects of this concept (what sort of mechanism? what friction? what happens when the mechanism is turned off while the vehicle is in motion? what effect on steering?). At this point, I am trying to understand the basic workings of the conceptual design. And, no, there's no intention of making a commercial product of this.

[herr_oberst]

It sounds like you're describing the perpetual motion machine..

[herr_oberst]

[javadog]

Originally Posted by jyl

Blah, blah, yammer, yammer, wheeze. Repeat a bunch, then stop

None of that is going to work, so stop thinking about it. Buy a good-quality cheeseburger for your upcoming lunch, wash it down with a quality beer and ponder the warm fuzzy feeling you get from that, for the rest of the day.

JR

[red-beard]

[LEAKYSEALS951]

Originally Posted by jyl

Use the drive mechanism to hold the weight at 3 o'clock. The reaction force on the wheel will rotate the wheel clockwise and propel the vehicle forward. Relative to the wheel, the weight will be rotating counterclockwise. Relative to the vehicle, the weight will be stationary.

I'm gonna jump in with a dumbarrse question. physics=20+yrs ago for me.
If the weight remains at 3 o'clock, it's potential energy is never translated into kinetic energy, therefore it is the imaginary mechanism holding the weight at 3 o'clock and spinning it which is powering the motion. Right? How is gravity relevant in this model?- unless the weight represents the maximum lifting force the imaginary mechanism can apply before the weight itself lifts relative to 3 o'clock

[aschen]

It sounds like you are hung up on the most fundamental bit, that a force applied at a velocity is power. (in scalar terms). It is a bit hard to rectify the first principle, but doesnt it follow your intuition?


If you are push a box along a floor that has 10lb friction force, doesn't it make sense that your rate of doing work (power) would be 2x as much at 10mph as it would be at 5mph for example? Or imagine riding a bike up hill, it seems intuitive that it takes a bunch more power to ride at a constant 100mph, vs say a 1 mph walking pace.


Your friends bicycle cheating device would work as hypothesized, but you will require more and more power for you mechanism to keep it at the required angle as the speed increases.

[aschen]

Originally Posted by LEAKYSEALS951

I'm gonna jump in with a dumbarrse question. physics=20+yrs ago for me.
If the weight remains at 3 o'clock, it's potential energy is never translated into kinetic energy, therefore it is the imaginary mechanism holding the weight at 3 o'clock and spinning it which is powering the motion. Right? How is gravity relevant in this model?- unless the weight represents the maximum lifting force the imaginary mechanism can apply before the weight itself lifts relative to 3 o'clock

The weight creates the torque on the wheel and weight is a force as a direct result of gravity

[Hendog]

Gererator?

[onewhippedpuppy]

You lost me after about one sentence. But one big thing stood out from all that - there is no such thing as "frictionless". Nope, period, not gonna happen. You can reduce friction, but nothing short of floating in the vacuum of space is frictionless.

[Hendog]

Once you have spent the energy of the weight descending to the bottom of the wheel you must replenish/inject energy to sustain the power. In your example you sustain your weight at the 3 o’clock position by attaching it to a pulley and hanging it from the frame of the “vehicle”: the two forces cancel each other out. If however you supply 20N of energy you should gain your acceleration (ignoring friction as you say).

[Pazuzu]

Originally Posted by LEAKYSEALS951

I'm gonna jump in with a dumbarrse question. physics=20+yrs ago for me.
If the weight remains at 3 o'clock, it's potential energy is never translated into kinetic energy, therefore it is the imaginary mechanism holding the weight at 3 o'clock and spinning it which is powering the motion. Right? How is gravity relevant in this model?- unless the weight represents the maximum lifting force the imaginary mechanism can apply before the weight itself lifts relative to 3 o'clock

You are correct, the weight won't do any work at all because there is no exchange of it's potential energy to kinetic energy.
The "mechanism" that is holding it is doing all of the work, so if the wheel rolls, you're having to put that energy into that mechanism somehow. Either winding a spring, or burning fuel or something.

If you don't do that, the wheel won't turn.