query162

course phy201

query 16.2 016. `query 16

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Question: `qClass notes #15

When a projectile rolls off a ramp with its velocity in the horizontal direction, why do we expect that its horizontal range `dx will be proportional to the square root of its vertical displacement `dy rolling down the ramp?

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Your solution:

Not sure

confidence rating #$&*

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Given Solution:

`a**

A quick synopsis:

The object accelerates uniformly downward, so the distance it falls is proportional to the square of the time of fall. Thus the time of fall is proportional to the square root of the distance fallen.

The object's horizontal velocity is constant, so its horizontal distance is proportional to the time of fall.

So the horizontal distance is proportional to the square root of the distance it falls.

More details:

The distance of vertical fall, starting with vertical velocity 0, is

`dy = v0 `dt + .5 a `dt^2 = 0 `dt + .5 a `dt^2 = .5 a `dt^2,

so `dy is proportional to `dt^2.

Equivalently, therefore, `dt is proportional to sqrt(`dy).

The horizontal distance is

`dx = v_horiz * `dt

so `dx is proportional to `dt.

`dx is proportional to `dt, and `dt is proportional to sqrt(`dy), so `dx is proportional to sqrt(`dy).

STUDENT RESPONSE WITH INSTRUCTOR COMMENTS:

When the

object is accelerating downward the distance it falls is equal to the square root of the time it takes to fall.

Right idea but that would be the square of the time, not the square root.

So the time of our fall is equal to proportional to, not equal to; again you have the right idea the square root of the distance the object fell. Our horizontal velocity is constant which will make our horizontal distance equal to the time of the fall. Because of this, the horizontal distance in equal to the square root of the distance that our object falls.

Good.

To summarize:

The vertical distance is proportional to the square of the time of fall, so the time of fall is proportional to the square root of the distance fallen.

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Self-critique (if necessary):

Ok, so

dy or (ds) = v0'dt +.5a 'dt^2 which is the vertical, V0=0 vertically so ds= .5a'dt

therefore dy or ds is proportional to 'dt^2 , so 'dt is proportional to sqrt dy.

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Question: `qIn the preceding situation why do we expect that the vertical kinetic energy of the ball will be proportional to `dy?

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Your solution:

Vertical KE is proportional to dy because 'dw= fnet * 'ds and 'dw = KE. Work done on the ball as it drops vertically are equal to the kinetic energy working on the ball as it drops.

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Given Solution:

`a**

This could be argued by analyzing the motion of the object, and using the definition of kinetic energy:

The vertical velocity attained by the ball is vf = `sqrt(v0^2 + 2 a `ds).

Since the initial vertical velocity is 0, for distance of fall `dy we have vf = `sqrt( 2 a `dy ), showing that the vertical velocity is proportional to the square root of the distance fallen.

Since KE is .5 m v^2, the KE will be proportional to the square of the velocity, hence to the square of the square root of `dy.

Thus KE is proportional to `dy. **

In terms of energy the argument is simpler:

PE loss is -m g `dy.

Since m and g are constant for this situation, PE loss is therefore proportional to `dy. (This means, for example, that if `dy is doubled then PE loss is doubled; if `dy is halved then PE loss is halved.)

KE gain is equal to the PE loss, so KE gain is also proportional to `dy.

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Self-critique (if necessary):

Ok,

so we have the vf^2= v0^2 + 2a 'ds. To find the vertical velocity which shows that vertical velocity is proportional to 'ds fallen. Once we have the vf we can use the KE= .5mv^2 which is proportional to the sqrtVf as well as sqrt Of dy (ds). proving that KE is proportional to dy (or ds vertically).

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Question: `qWhy do we expect that the KE of the ball will in fact be less than the PE change of the ball?

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Your solution:

I am not sure hree, I thought that KE is equal to PE loss, so when KE is positive Pe would be negative but opposite to the vlaue of KE.

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Given Solution:

`a** There is some air friction, which dissipates some of the energy. PE is lost and the lost PE goes into an increase in KE, and into dissipated energy. The KE increase and dissipated energy 'share' the 'lost' PE.

STUDENT RESPONSE: Because actually some of the energy will be dissipated in the rotation of the ball as it drops?

INSTRUCTOR COMMENT: Good try. However there is KE in the rotation, so rotation accounts for some of the KE but doesn't dissipate KE. Rotational KE is recoverable--for example if you place a spinning ball on an incline the spin can carry the ball a ways up the incline, doing work in the process. Dissipated energy is not recoverable.

The PE loss is converted to KE, some into rotational KE which doesn't contribute to the range of the ball and some of which simply makes the ball spin.

ANOTHER STUDENT RESPONSE: And also the loss of energy due to friction and conversion to thermal energy.

INSTRUCTOR COMMENT: Good. There would be a slight amount of air friction and this would dissipate energy as you describe here, as would friction with the ramp (which would indeed result in dissipation in the form of thermal energy). **

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Self-critique (if necessary):

Ok, so air friction, dissipating energy, lowers the PE and Increases the KE.

??????????????????????So does this mean that they do not neccessarily have to be equal and opposite of each other??????????????????????

Good questions.

Check out this explanation, which was added to the original document:

Air friction doesn't lower the PE; it's

the change in altitude that lowers the PE (an object's gravitational potential

energy gets lower as it descends).  The loss of PE results in an increase

in KE, though not as great an increase as if air friction wasn't present.

Related to the idea that `dKE and `dPE

should be equal and opposite (which is sometimes the case and sometimes not):

The energy situation is governed by

the work-energy theorem in the form

In general, a nonconservative force

can increase the KE and/or the PE in any way at all. It can increase one

without changing the other. `dKE and `dPE are not generally equal and

opposite. For example you can speed up a cart by pushing or pulling it along

a level surface. The force you exert is nonconservative, and it increases

the KE of the cart without changing its PE.  Or you could lift an

object at a constant speed; its KE wouldn't change but its gravitational PE

would.


However in some situations nonconservative forces are either absent or

negligible. For example if you toss a steel ball a couple of meters into the

air and let it fall to the ground, it doesn't attain enough speed for air

resistance to become significant, so once you release it the ball pretty

much behaves as if nonconservative forces were absent. Then as it rises it

slows down, decreasing its KE and increasing its PE. As it then falls its PE

decreases but it speeds up, increasing its KE. Changes in KE and PE turn

out, in this case, to be equal and opposite.