| Topic: |
Science > Physics |
| User: |
"mutedHyperbole" |
| Date: |
14 Nov 2006 05:10:12 PM |
| Object: |
Kinetic energy of a wheeled-vehicle |
A car weighs C Kgs.
C = A + B
Where A is the mass of the wheels (all 4 of them), and B is the mass of
the car without the wheels.
Is the total kinetic energy of the moving car equal to:
1. 0.5*C*V^2
or is it...
2. 0.5*B*V^2 + 2*(1/2*I*w^2) --> I is the inertia of the wheels, and w
is the angular rotational speed.
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| User: "CWatters" |
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| Title: Re: Kinetic energy of a wheeled-vehicle |
14 Nov 2006 05:36:40 PM |
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"mutedHyperbole" <zutalors212@yahoo.com> wrote in message
news:1163545812.257606.41300@f16g2000cwb.googlegroups.com...
A car weighs C Kgs.
C = A + B
Where A is the mass of the wheels (all 4 of them), and B is the mass of
the car without the wheels.
Is the total kinetic energy of the moving car equal to:
1. 0.5*C*V^2
or is it...
2. 0.5*B*V^2 + 2*(1/2*I*w^2) --> I is the inertia of the wheels, and w
is the angular rotational speed.
Did you mean
0.5*B*V^2 + 4*(1/2*I*w^2)
as there are four wheels not two?
Anyway isn't it...
0.5*C*V^2 + 4*(1/2*I*w^2)
The wheels have BOTH velocity and rotation (so C rather than B).
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| User: "" |
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| Title: Re: Kinetic energy of a wheeled-vehicle |
14 Nov 2006 09:34:48 PM |
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CWatter, you have sharp eyes and an even sharper mind.
The total kinetic energy of the vehicle is the combination of Mv^2 +
4(Iw^2) where M is the combined mass of the vehicle and wheel weight.
Curiously, this is a question on the "Oxford University" problem set
which is included in the homework exercises of a number of Americal
texts on classical mechanics. This makes me inclined to believe that
the original post was from an undergraduate college student search for
help on his homework.
To another poster that commented in this thread, realize that the total
energy expended on impact depends on the coefficient of friction of
wheel/ground interface. Still, that was not the question posed.
Harry C.
CWatters wrote:
"mutedHyperbole" <zutalors212@yahoo.com> wrote in message
news:1163545812.257606.41300@f16g2000cwb.googlegroups.com...
A car weighs C Kgs.
C = A + B
Where A is the mass of the wheels (all 4 of them), and B is the mass of
the car without the wheels.
Is the total kinetic energy of the moving car equal to:
1. 0.5*C*V^2
or is it...
2. 0.5*B*V^2 + 2*(1/2*I*w^2) --> I is the inertia of the wheels, and w
is the angular rotational speed.
Did you mean
0.5*B*V^2 + 4*(1/2*I*w^2)
as there are four wheels not two?
Anyway isn't it...
0.5*C*V^2 + 4*(1/2*I*w^2)
The wheels have BOTH velocity and rotation (so C rather than B).
.
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| User: "" |
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| Title: Re: Kinetic energy of a wheeled-vehicle |
14 Nov 2006 05:35:01 PM |
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mutedHyperbole <zutalors212@yahoo.com> wrote:
A car weighs C Kgs.
C = A + B
Where A is the mass of the wheels (all 4 of them), and B is the mass of
the car without the wheels.
Is the total kinetic energy of the moving car equal to:
1. 0.5*C*V^2
or is it...
2. 0.5*B*V^2 + 2*(1/2*I*w^2) --> I is the inertia of the wheels, and w
is the angular rotational speed.
Suppose the car is REALLY strong, going at some speed, flys off a
cliff, and is totally stopped by a corresponding REALLY strong wall.
Are the wheels still turning until the car hits the ground?
Is the energy in the spinning wheels kinetic energy or something else?
--
Jim Pennino
Remove .spam.sux to reply.
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| User: "Igor" |
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| Title: Re: Kinetic energy of a wheeled-vehicle |
15 Nov 2006 11:34:21 AM |
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mutedHyperbole wrote:
A car weighs C Kgs.
C = A + B
Where A is the mass of the wheels (all 4 of them), and B is the mass of
the car without the wheels.
Is the total kinetic energy of the moving car equal to:
1. 0.5*C*V^2
or is it...
2. 0.5*B*V^2 + 2*(1/2*I*w^2) --> I is the inertia of the wheels, and w
is the angular rotational speed.
The combined kinetic energy will always be the translational kinetic
energy of the center of mass, which includes the mass of the whole car,
wheels and all, plus the rotational kinetic energy of the wheels. Thus
the total kinetic energy would be 0.5 C V^2 + 2 I w^2.
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