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course phy201
In today's class we encountered the following definitions:The line of a force is the straight line parallel to the force which passes through the point at which the force is applied.
The moment arm of a force relative to an axis is the vector from the axis to the line of force.
The torque `tau produced by a force `F with moment arm `r has magnitude
tau = r * F
The torque `tau is positive if the force tends to produce counterclockwise rotation, negative if it tends to produce clockwise rotation. (More generally the torque is the cross product `r X `F. University physics students will understand this. Others are unlikely to be familiar with the concept, which will be developed when and if needed).
Newton's Second Law applied to rotational systems is
`tau_net = I * `alpha
where alpha is the angular acceleration of the system.
The work done by torque `tau as a system rotates through angular displacement `dTheta is
`dW = `tau * `dTheta
(qualifier for University Physics students: this is so if `tau and `dTheta are parallel; in general the relationship is `dW = (component of `tau in the direction of `dTheta) * `dTheta, or more concisely `dW = `tau dot `dTheta).
The rotational kinetic energy of a rotating system is
KE = 1/2 I omega^2.
We reasoned out the torque produced by a given force applied at a given point perpendicular to a rod of given mass and length, rotating about its center. We also found its moment of inertia, which (assuming that torque to be the net torque) allowed us to calculate its acceleration. Then given its angular displacement in a given time interval, we reasoned out its average angular velocity, then its change in angular velocity, then its initial and final angular velocities.
It was then suggested that everyone work out the corresponding angular displacement and the word done by the torque, assuming it to be the net torque. The initial and final kinetic energies, and then the change in kinetic energy, are easily calculated. The change in kinetic energy should then be compared with the work done by the net torque.
Having made all these comparisons, and having reasoned out the units of all quantities, the entire idea of rotational motion and its analysis should begin to 'click'.
`q001. Report your data for today's experiment, including a brief explanation of what you did and how you made your measurements.
15 g dominoes with 10 dominoes in a bag, which stretches 24cm, and with 5 dominoes it stetches 18 cm
The 10 dominoes oscillated at 2.2 oscill/sec````````9.04 rad/s
5 dominoes oscillated at 1.4 oscill/sec``````````12.78 rad/s
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I think you mixed up your numbers: 2.2 osc / sec = 12.8 rad/s and 1.4 osc / sec = 9.04 rad / sec.
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`q002. What was the slope of your force vs. length graph for the rubber band chain?
12.25 N/m
For each oscillating system you measured, how consistent was the value of omega as determined from the graph slope and the suspended mass with the value of omega as determined from timing oscillations?
consistent
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You need to show this. For your system I get very roughly 8 rad/s and 11 rad / s, which is pretty close to the observed angular frequencies.
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`q003. A point moves around a reference circle with angular velocity 2 radians / second. At t = 0 the point is at the positive x axis.
What is the angular position of the point at t = 1 second, t = 2 seconds and t = 3 seconds?
1 sec-2 rad
2 sec- 4 rad
3 sec- 6 rad
How long does it take the point to go once around the circle?
2x2pi
12.56 s
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2 * 2 pi has no meaning associated with this situation, nor does 12.56 s.
You show that the point goes through 6 radians in 3 seconds. 6 radians is pretty close to a full circle. So it takes a little over 3 seconds.
How would you calculate this more precisely?
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When is the time t at which the point first reaches the angular position 5 radians, in which quadrant is it when this occurs, and at that instant is it closer to the x or the y axis?
1- Near the y axis
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I agree, but you haven't yet specified the time at which this occurs.
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How many times will it have gone around the circle, in what quadrant will it be, and will it be closer to the x or y axis after 8 seconds have elapsed?
¾, quad 3 or 4
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As you showed earlier, after 3 seconds the angular position will be 6 radians, very nearly a full circle. So in 8 seconds the point is going to go through more that two complete circles.
You can determine for certain in which quadrant it will be at that time.
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`q004. This question is much like the preceding except we're going to use symbols.
A point moves around a reference circle with angular velocity omega. At t = 0 the point is at the positive x axis.
What is the angular position of the point at clock time t?
Anf vel/ clock time = tposition
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This is not consistent with the definition of average angular velocity.
What is that definition and how is it related to the answer to this question?
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How long does it take the point to go around the circle?
T position x 2pi
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The angular displacement, which is the change in angular position, around the complete circle is 2 pi radians. The angular velocity is omega.
How does the definition of angular velocity as the rate of change of angular position with respect to clock time relate change in clock time, change in angular position and angular velocity? What is the specific relationship and how can it be used to answer this question?
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What is the clock time when the point first reaches the negative y axis?
¾ of the total
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That would be correct. If you have the expression for the total time required, the answer to this question would be 3/4 of that.
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Let theta_1 stand for some angular position between 0 and 2 pi radians. At what clock time does the point first reach this angular position?
Thetapos/velocity=clocktime
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From the definition you could conclude that the change in angular position divided by the angular velocity is equal to the change in clock time. You could have stated it more clearly (it's change in clock time, not clock time) but you have the right idea.
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What is the clock time when the reference point again reaches this angular position?
Clocktime+ 1 full circle(2pi)
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using the quantity you called clock time, adding this to the time required to go around the full circle you would get the correct answer. The expression for the time to go around the full circle is not 1 full circle ( 2 pi ).
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`q005. Let the point again move around the circle at 2 radians / second, and assume the circle to have radius 10 cm.
What is the x coordinate of its position at t = 0.2 seconds, 0.4 seconds and 0.6 seconds?
.4 rad
.8 rad
1.2 rad
How long does it take before the x coordinate of the point becomes negative?
At pi
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The x coordinate becomes negative when the angular position passes the positive y axis, which occurs at angular position pi/2.
At 2 pi rad / sec, how long would this take?
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How fast is the point moving?
20 cm/s
What is its centripetal acceleration?
40 cm/s^2
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good
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When its angular position is 60 degrees, what is the direction of its velocity vector, as measured counterclockwise relative to the positive x direction?
Positive y axis
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The velocity vector is parallel to the y axis when it is vertical, which occurs when the point is at the positive or negative x axis. This does not occur when the angular position is 60 degrees.
You need to have a decent picture of the 60 degree position and the velocity vector, which is tangent to the circle (and which therefore makes angle 90 degrees with the radial vector, that is the vector from the origin to the reference point).
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When its angular position is 60 degrees, what is the direction of its centripetal acceleration vector, as measured counterclockwise relative to the positive x direction?
Towards the origin
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It is toward the origin, but what therefore is its angle with the positive x direction?
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When its angular position is 60 degrees, what are its x and y coordinates?
Pos. x,y
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If the vector from the origin to the point is R, and the angle of that vector with the positive x axis is theta, then by the basic vector relationships the x and y components are
R_x = R cos(theta)
R_y = R sin(theta).
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When its angular position is 60 degrees, what are the x and y coordinates of its velocity vector?
Pos x pos y
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If v is the velocity and theta the angle of the velocity vector (which will be different than the theta for the radial vector) then by the usual relationships you will have
v_x = v cos(theta)
v_y = v sin(theta).
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Check my notes and resubmit what you can.
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