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course phy 201
`q001. Report your data for today's experiment, including a brief explanation of what you did and how you made your measurements.We used 15g dominos and measured how far 10 and 5 dominos would stretch a rubber band when they were in a sandwich bag. 10 domino stretched the bag 24 cm and ossolated 2.2 ossolations/s and the omega was found to be 9.04 radians/s. 5 dominos stretched the rb 18cm and ossolated 1.4 ossolations/s and omega was 12.78 radians/s.
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`q002. What was the slope of your force vs. length graph for the rubber band chain?
12.25 N/m
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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?
consistant
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You need to show your results and how you got them. You also need to address the degree of consistency.
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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?
1sec = 2radians
2sec = 4radians
3sec = 6radians
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How long does it take the point to go once around the circle?
2*2pi = 12.56s
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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?
Near the yaxis
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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?
.75 of the way around the circle
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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?
Angular v/time = t position
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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?
Tposition * 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?
Theta pos/velocity = clock time
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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?
Im not sure how to calculate this
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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?
.2s = 0.4 radians
.4s = .8 radians
.6s = 1.2radians
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How long does it take before the x coordinate of the point becomes negative?
Pi(r)
2rad/s
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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?
20cm/s
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What is its centripetal acceleration?
V^2/r
A centripetal = 40 cm/s^2
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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?
Pos 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 and pos y axis
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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?
X= cos(60)
Y = sin(60)
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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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When its angular position is 60 degrees, what are the x and y coordinates of its centripetal acceleration vector?
As it gets closer to the center it is going to increase speed
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at 90 deg, cosine is 0 and decreasing, sine is 1 and decreasing
at 0 deg, cosine is 1 and decreasing (just like sine at 90 deg), sine is 0 and increasing (just like cosine at 270 deg)
so
cos(alpha) = sin(alpha + 90 deg)
sin(alpha) = cos(alpha + 270 deg) = cos(alpha - 90 deg)
not formulas to be memorized, or tables to be referenced, but something we can always reason out if we think clearly
where do values of the sine act like the values of the cosine act here
where do values of the cosine act like the values of the sine act here
**** The sin and cos do the same thing one is just for the x and the other is for the y adding the 90 degrees just makes it to where it is positive instead of negative.
cos(90 deg + alpha) = -sin(alpha)
sin(90 deg + alpha) = cos(alpha)
etc. for 180 and 270 deg
also for pi2, pi, 3 pi/2
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Check my notes and resubmit what you can.
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