10-15

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course Phy201

`q001. A domino of mass .02 kg is moving at 140 cm / second. What is its kinetic energy?KE = 1/2mv^2

KE = ½(.02kg)(1.4m/s)^2

KE = 0.02J

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If the kinetic energy of a domino of mass .02 kg is .5 Joules, how fast is it moving?

KE = 1/2mv^2

.5J = ½ (0.02kg)(v)^2

V = 7.02m/s

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An object of unknown mass is given a kinetic energy of 8 Joules, which causes it to move at 12 meters / second. What is its mass?

KE = 1/2mv^2

8= ½(m)(12m/s)^2

M = 0.03 kg

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All the above questions would be answered using a single definition. What is that definition?

KE = 1/2mv^2

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`q002. A domino of mass .02 kg is positioned on a rotating strap, 20 cm from the axis of rotation. How far does the domino move as the strap completes a full rotation, and how far does it move as the strap rotates through 220 degrees?

Im not sure how to attempt this question when given in degrees. What formulas are used to determine how far the object moves. ?????

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@&
Once you understand that the path of the object is a circle, you solve this using the basic geometry of a circle.
*@

How fast is the domino moving if the strap is rotating at 220 degrees / second, and what is its kinetic energy at that speed?

****

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Through how many degrees per second would the strap have to be rotating in order for the domino to have 0.1 Joules of kinetic energy?

****

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These questions would be answered with the same definition used for the first question, in addition to what common knowledge from high school mathematics?

****

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`q003. A horse of mass 500 kg starts from rest and descends a steep bank, losing 50 000 Joules of gravitational potential energy. How fast would the horse be moving if all the lost potential energy was converted to kinetic energy?

PE = -dKE

KE = 50,000J

KE = 1/2mv^2

50,000 = ½(500)v^2

Sqrt 200 = sqrt v^2

V = 14.1 m/s

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If at the bottom of the bank the horse in the preceding is moving at 10 meters / second, how much work was done on it by nonconservative forces?

KE = 1/2mv^2

KE = ½(500kg)(10m/s)^2

KE = 25,000J

Wcons = (500kg)(9.8 m/s^2)= 4,900J

@&
This calculation has units of kg m/s^2, not Joules. This reveals that something is amiss. Basically, you've calculated something important, but it's not work.
*@

25,000J - 4,900J = 20,100J

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@&
Right idea, but the work is not 4900 Joules.
*@

What might be the nature of these nonconservative forces?

Friction and air resistance

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The first two questions were answered based on what definitions and what theorems?

KE = 1/2mv^2 and dW = dKE and the work energy theorem

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`q003. What is the kinetic energy of a rotating system whose moment of inertia is .012 kg m^2 when it is rotating at 4 radians / second?

KE = ½(I)(omega)^2

KE = ½ (0.012 kgm^2)(omega)^2

KE = 0.096J

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What is the angular velocity of this rotating system if its kinetic energy is 0.6 Joules?

0.6J = ½(0.012kgm^2)(omega)^2

Sqrt 100 = sqrt omega ^2

Omega = 10m/s

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@&
If you use the units you don't end up with m/s. The meters all divide out.

You end up with 1/s (reciprocal seconds), which can be interpreted as radians / second.
*@

What is the moment of inertia of a rotating system whose kinetic energy is 500 Joules when it is rotating at 40 radians / second?

KE = ½(I)(omega)^2

500J = ½(I)40m/s)^2

I = 0.625 kgm^2

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What definitions were required to answer these questions?

KE = 1/2mv^2

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@&
You used KE = 1/2 I omega^2.
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`q004. What is the moment of inertia of a styrofoam wheel in which 12 bolts, each of mass 100 grams, are inserted around a circle 50 cm in diameter and concentric with the axis of rotation, with another 8 bolts, each of mass 150 grams, inserted around a concentric circle 30 cm in diameter? Ignore the moment of inertia of the styrofoam itself.

I = mr^2

I = 1.2kg(0.25m)^2

I = 0.075kgm^2

I = 1.2kg(0.15m)^2

I = 0.027kgm^2

Is it necessesary to combine the two systems????

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@&
All those masses are rotating, and if you want to speed them up you have to speed them all up.

So you would add the moments of inertia.
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`q005. A long metal spring exerts forces of 500 Newtons, 400 Newtons, 250 Newtons, 100 Newtons, and 5 Newtons at respective lengths 20 meters, 18 meters, 16 meters, 13 meters and 10 meters.

What is its average tension between the 16 meter length and the 13 meter length?

About 179N

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How much work would be required to stretch the cord from the 13 meter length to the 16 meter length?

W = F*ds

W = 179N * 3m = 537J

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If the tension is conservative, how much potential energy would be lost as the cord's length decreases from 16 meters to 13 meters?

PE = -537J

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Assuming the tension to be conservative, how much potential energy is built on each of the four intervals, as the spring is stretched from its 10 meter length to its 20 meter length?

PE = -(179N)(10m) = -1790J

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@&
You wouldn't use the average force between 16 m and 13 m to calculate the work from 20 m to 10 m. You have the information for three length intervals, so you should use it to get a reasonably accurate estimate.
*@

If all that potential energy could be converted to the kinetic energy of a rotating system whose moment of inertia is 2 kg m^2, what would be the angular velocity of that system?

KE = ½(I)(omega)^2

1790 = ½(2kgm^2)(omega)^2

Sqrt 1790 = sqrt omega^2

Omega = 42.3m/s

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What definitions are required to answer these questions?

KE = ½(I)(omega)^2 and PE = -KE and work energy theorem

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@&
See my previous note; if units are calculated correctly there are no meters in this result.
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`q006. How long would it take a freely falling object, starting from rest, to reach a speed of 300 meters / second, and how far would it travel during this time?

Vf = 300m/s

Vo = 0m/s

Vave = 150m/s

dt = 2s

ds = 300m

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What would be the change in the gravitational potential energy of a 100 kg mass falling in the manner described here?

W = F*ds

F = ma

F = (100kg)(75m)

@&
The acceleration of a freely falling object is not 75 m.

75 m is not an acceleration in any case.
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F = 7500 N

@&
The units of your calculation would be kg * m, not Newtons.

Had you used an acceleration you would have obtained units of kg m/s^2, or Newtons.
*@

W = 7500N (300m)

W = 2,250,000

dw = dKE

dPE = -KE

PE = -2,250,000J

This seems like an awful big number ??

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@&
When you correct your force, the number will still be big, but not quite that big.
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If due to air resistance a 100 kg mass, falling from rest, requires 40 seconds to reach a speed of 300 meters / second, then what is your best estimate of the average force of air resistance?

F = ma

F = (100kg)(300m/40s)

F = 750N

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@&
300 m / (40 s) is not an acceleration, and the units of the calculation you show would be kg m/s, not Newtons.
*@

@&
A lot of your reasoning is good, but you'll want to make some modifications.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.

10-15

@& Once you understand that the path of the object is a circle, you solve this using the basic geometry of a circle. *@

@& This calculation has units of kg m/s^2, not Joules. This reveals that something is amiss. Basically, you've calculated something important, but it's not work. *@

@& Right idea, but the work is not 4900 Joules. *@

@& If you use the units you don't end up with m/s. The meters all divide out. You end up with 1/s (reciprocal seconds), which can be interpreted as radians / second. *@

@& You used KE = 1/2 I omega^2. *@

@& All those masses are rotating, and if you want to speed them up you have to speed them all up. So you would add the moments of inertia. *@

@& You wouldn't use the average force between 16 m and 13 m to calculate the work from 20 m to 10 m. You have the information for three length intervals, so you should use it to get a reasonably accurate estimate. *@

@& See my previous note; if units are calculated correctly there are no meters in this result. *@

@& The acceleration of a freely falling object is not 75 m. 75 m is not an acceleration in any case. *@

@& The units of your calculation would be kg * m, not Newtons. Had you used an acceleration you would have obtained units of kg m/s^2, or Newtons. *@

@& When you correct your force, the number will still be big, but not quite that big. *@

@& 300 m / (40 s) is not an acceleration, and the units of the calculation you show would be kg m/s, not Newtons. *@

@& A lot of your reasoning is good, but you'll want to make some modifications.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).Be sure to include the entire document, including my notes.

@&
Once you understand that the path of the object is a circle, you solve this using the basic geometry of a circle.
*@

@&
This calculation has units of kg m/s^2, not Joules. This reveals that something is amiss. Basically, you've calculated something important, but it's not work.
*@

@&
Right idea, but the work is not 4900 Joules.
*@

@&
If you use the units you don't end up with m/s. The meters all divide out.

You end up with 1/s (reciprocal seconds), which can be interpreted as radians / second.
*@

@&
You used KE = 1/2 I omega^2.
*@

@&
All those masses are rotating, and if you want to speed them up you have to speed them all up.

So you would add the moments of inertia.
*@

@&
You wouldn't use the average force between 16 m and 13 m to calculate the work from 20 m to 10 m. You have the information for three length intervals, so you should use it to get a reasonably accurate estimate.
*@

@&
See my previous note; if units are calculated correctly there are no meters in this result.
*@

@&
The acceleration of a freely falling object is not 75 m.

75 m is not an acceleration in any case.
*@

@&
The units of your calculation would be kg * m, not Newtons.

Had you used an acceleration you would have obtained units of kg m/s^2, or Newtons.
*@

@&
When you correct your force, the number will still be big, but not quite that big.
*@

@&
300 m / (40 s) is not an acceleration, and the units of the calculation you show would be kg m/s, not Newtons.
*@

@&
A lot of your reasoning is good, but you'll want to make some modifications.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.