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Phy 241
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** CQ_1_19.2_labelMessages **
Sketch a vector representing a 10 Newton force which acts vertically downward.
Position an x-y coordinate plane so that the initial point of your vector is at the origin, and the angle of the vector as measured counterclockwise
from the positive x axis is 250 degrees. This will require that you 'rotate' the x-y coordinate plane from its traditional horizontal-vertical orientation.
answer/question/discussion: ->->->->->->->->->->->-> :
I sketched a vector in the vertical downward position with a magnitude of 10. I placed a x-y coordinate system at the origin of the vector a rotated just slightly
so that the vector is at 250° counterclockwise from the positive x-axis
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What are the x and y components of the equilibrant of the force?
answer/question/discussion: ->->->->->->->->->->->-> :
the x and y components of the equilibrant of the force are the opposite of the x and y components of the force.
x = 10sin(20) = 3.4
y = 10cos(20) = 9.4
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&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
`R = -cos t `i + sin t `j.
`u_r is the unit vector in the direction of `R. Since `R is a unit vector, `u_r = `R = -cos t `i + sin t `j.
We can also express `u_r in terms of theta, where theta is expressed as a function of t.
It should be clear that as t goes from 0 to pi/2 to pi to 3 pi/2 to 2 pi, the trace of `R goes around the unit circle in the negative direction, starting at (-1, 0) and passing through (0, 1), (1, 0), (0, -1) before returning to (-1, 0). Thus as t goes from 0 to 2 pi, the angular position theta goes from pi to pi/2 to 0 to -pi/2 to -pi. From this we can conclude that theta(t) = pi - t. (It's also worth visualizing how this function and `R(t) = cos(t) `i + sin(t) `j describe two points moving around the circle in opposite directions, with the two points always symmetric about the y axis).
More formally we see that for our function R(t), the angular position at t is
theta = arcTan( sin(t) / (-cos(t) ) = arcTan(-tan(t)) + pi = arcTan(tan(-t)) + pi = -t + pi.
Again, theta = pi - t
Either way, we have
`R(t) = `R(theta(t)) = cos(theta(t)) `i + sin(theta(t)) `j = cos(pi-t) `i + sin(pi-t) `j.
Now
`u_r = cos(theta) `i + sin(theta) `j
and
`u_theta = -sin(theta) `i + cos(theta) `j.
`u_r ' = u_theta
and
`u_theta ' = - u_r.
`u_r = cos(pi - t) `i + sin(pi - t) `j
`u_theta = -sin(pi-t) `i + cos(pi - t `j)
`R = `u_r
`R ' = `u_r ' = sin(pi - t) `i - cos(pi-t)`j = - `u_theta
`R '' = cos(t) `i - sin(t) `j = -`R
`R = cos(theta) `i + sin(theta) `j
`R ' = u_theta theta ' = -u_theta
`R '' = -u_theta ' = -(-u_r) theta ' = -u_r(theta)
theta = arcTan(sin(t) / (-cos(t) ) = arcTan(-tan(t)) = arcTan(tan(-t)) + pi = pi - t.
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`R = -cos t `i + sin t `j.
`u_r is the unit vector in the direction of `R. Since `R is a unit vector, `u_r = `R = -cos t `i + sin t `j.
We can also express `u_r in terms of theta, where theta is expressed as a function of t.
It should be clear that as t goes from 0 to pi/2 to pi to 3 pi/2 to 2 pi, the trace of `R goes around the unit circle in the negative direction, starting at (-1, 0) and passing through (0, 1), (1, 0), (0, -1) before returning to (-1, 0). Thus as t goes from 0 to 2 pi, the angular position theta goes from pi to pi/2 to 0 to -pi/2 to -pi. From this we can conclude that theta(t) = pi - t. (It's also worth visualizing how this function and `R(t) = cos(t) `i + sin(t) `j describe two points moving around the circle in opposite directions, with the two points always symmetric about the y axis).
More formally we see that for our function R(t), the angular position at t is
theta = arcTan( sin(t) / (-cos(t) ) = arcTan(-tan(t)) + pi = arcTan(tan(-t)) + pi = -t + pi.
Again,
theta = pi - t
Either way, we have
`R(t) = `R(theta(t)) = cos(theta(t)) `i + sin(theta(t)) `j = cos(pi-t) `i + sin(pi-t) `j.
Now it still holds, as always, that
`u_r = cos(theta) `i + sin(theta) `j
and
`u_theta = -sin(theta) `i + cos(theta) `j.
If ' stands for the derivative with respect to theta we have, as always
`u_r ' = u_theta
and
`u_theta ' = - u_r.
Remember, we are using ' for the derivative with respect to theta. To get the derivatives of `u_r and `u_theta with respect to t, since theta is a function of t we need to apply the chain rule. We can calculate the derivatives as follows:
d ( `u_r(theta) ) / dt = d(theta) / dt * `u_r ' = -`u_theta (theta)
and
d( `u_theta (theta) ) / dt = d(theta) / dt * `u_theta ' = - (-`u_r (theta)) = `u_r (theta).
`u_r = cos(pi - t) `i + sin(pi - t) `j
`u_theta = -sin(pi-t) `i + cos(pi - t `j)
`R = `u_r
`R ' = `u_r ' = sin(pi - t) `i - cos(pi-t)`j = - `u_theta
`R '' = cos(t) `i - sin(t) `j = -`R
`R = cos(theta) `i + sin(theta) `j
`R ' = u_theta theta ' = -u_theta
`R '' = -u_theta ' = -(-u_r) theta ' = u_r(theta)
theta = arcTan(sin(t) / (-cos(t) ) = arcTan(-tan(t)) = arcTan(tan(-t)) + pi = pi - t.
We can now answer the given question in two ways.
FIrst, we find the velocity and acceleration vectors directly:
`R = -cos t `i + sin t `j
so
`R ' = sin(t) `i + cos(t) `j
and
`R '' = cos(t) `i - sin(t) `j.
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&#Good responses. Let me know if you have questions. &#