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PHY 201
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I am submitting a few trial test questions to see if I am on the right track.
Problem Number 3
A uniform disk of mass .21 kg and radius 10 cm is constrained to rotate on an axis about its center. Friction exerts a net torque of .00126 meter Newtons on the system when it is in motion. On the disk are mounted masses of 9 grams at a distance of 8.5 cm from the center, 8 grams data distance of 5.7 cm from the center and 45 grams at a distance of 3.2 cm from the center. A uniform force of .09 Newtons is applied at the rim of the disk in a direction tangent to the disk. The force is applied for 9 seconds with the disk initially at rest.
• Using energy considerations find the KE attained by the system.
• At the system's final angular velocity what is the speed and KE of each of the masses mounted on the disk?
Mass 1- 9 grams and 8.5 cm, mass 2—8 g, 5.7 cm, mass 3—45 grams at 3.2 cm.
Omega=2pi(.1m)*9 sec=5.65 m/s
m * sec doesn't give you m / s
omega is rate of change of angular position with respect to clock time, so omegaAve = (change in angular position) / (change in clock time).
The change in angular position could be measured in revolutions, radians, degrees, half-revolutions, etc..
However for these applications, radians are the most convenient measure.
Thus omega is measured in radians / second.
Inertia=mr^2
That's 'moment of inertia'; the unmodified word 'inertia' has a different meaning. To avoid possible confusion it's best to use the correct terminology.
Disk=.21 kg* (.1m)^2=.0021 kg m^2 the moment of inertia of a disk is 1/2 M R^2
Mass 1=.009 kg*.085 m^2=.6.5 x10^-5 kg m^2
Mass 2=.008 kg*.057 m^2=2.6 x10^-5 kg m^2
Mass 3=.045 kg *.032 m^2=4.6x10^-5 kg m^2
Sum of inertia=.00224 kg m^2
the moment of inertia of the disk is only about 3.3 * 10^-5 kg m^2, so the total moment of inertia will only be about .0019 kg m^2.
KE=1/2 (.00224 kg m^2)*5.65 m/s^2=0.036 N of the system
kg * m^2 * m / s^2 = kg * m^3 / s^2; N is kg m/s^2 so the units of your calculation here would be N * m^2.
However the units of omega are rad / sec, so the correct units would be kg * m^2 * rad / s^2. Since m * rad = m (a meter of radius times a radian of angle gives you a meter of arc distance), you get simply kg m^2 / s^2, which is N * m or Joules.
5.65 isn't the right number, though (see my note on the calculation that gave you this number).
To solve the second question; you’d solve for omega at each mass and then use the inertia to find the speed and KE. OR could you use v=omega*r to find the speed at each spot.
The key thing you are missing is that the KE attained by the system is the work done by the net force.
In this case the .09 N force, exerted at distance 10 cm from the axis, exerts a torque of .09 N * .1 m = .009 m * N.
Friction exerts a torque of .00126 m N in the opposite direction.
So the net torque is .0077 m N in the direction of the .009 m * N torque.
Just as force * `dt = change in momentum = m `dv (force * `dt = impulse), torque * `dt = change in angular momentum = I `dOmega. So
.0077 m N * 9 sec = .0019 kg m^2 * `dOmega so that
`dOmega = .0077 m N * 9 sec / (.0019 kg m^2) = 36 rad / sec, approx..
Problem Number 5
A simple pendululm of length 2 meters and mass .45 kg is pulled back a distance of .07 meters in the horizontal direction from its equilibrium position, which also raises it slightly. How much force tends to pull the pendulum back to its equilibrium position at this point?
Force pulling back=mg*`dx=.45 kg *9.8 m/s^2*.07 m=.3087 N
kg * m/s^2 * m = kg m^2 / s^2 = Joules, or maybe m * N, but not Newtons
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That would be m g / L * `dx. L is about 2 m. That fixes the units problem as well, and you get about .16 N.
The reasoning behind this result: The string is nearly vertical so the tension is nearly equal to m g. The horizontal displacement is in proportion `dx / L to the length of the pendulum, and the horizontal component of the force is in the same proportion to the tension, and directed opposite to the displacement. So the horizontal tension is T * (-`dx) / L = = m g * `dx / L, which we rewrite as -(m g / L) * `dx in order to put it in the same form as -k x.
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AND the hill question:
Problem Number 4
Coasting from rest down a certain hill, whose slope is variable, I reach a speed of 10 m/s at the halfway point. The total height of the hill is 2.891 meters. Ignoring the effects of air resistance and friction:
• How fast would I therefore be going if I coasted down the second half of the hill from rest?
.I am still confused on how you translate a speed of 10 m/s at halfway and a total height of 2.891 meters into PE
`dPEac=`dPEab+`dPEbc If I know a halfway speed of 10 m/s and the height of 1.455m, and the height of the second half to be 1.455 m; convert into `dPE.
Write out your expressions for KE and `dPE.
Let M stand for the mass of the car. Then at 10 m/s its KE is 1/2 M * (10 m/s)^2 = M * 50 m^2 / s^2.
You don't know anything about slope so you don't know anything about the vertical coordinate of the car when it's halfway down. You can't assume that both halves have the same average slope.
You do know that the car loses M g * 2.9 meters of PE, equal to M * 9.8 m/s^2 * 2.9 m = M * 28 m^2 / s^2.
So if A, B and C are the positions at the top, halfway point, and bottom, we know that
`dPE_AC = -M * 28 m^2 / s^2
`dKE_AB = + M * 50 m^2 / s^2.
It follows that `dPE_AB = - M * 50 m^2 / s^2.
Now `dPE_AC = `dPE_AB + `dPE_BC, so we can find `dPE_BC:
`dPE_BC = `dPE_AC - `dPE_AB = - M * 28 m^2 / s^2 - (-M * 50 m^2 / s^2) = + M * 22 m^2 / s^2.
`dPE_BC = M g `dy_BC = M * 22 m^2 / s^2, so
`dy_BC = 2.2 meters.
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