Show that the radial acceleration is zero when θ = 1/sqrt(2) radians
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Show that the radial acceleration is zero when θ = 1/sqrt(2) radians
Transcribed Image Text:A particle moves outward along a spiral. Its trajectory is given by r = A0, where A is a constant.
A = (1/n) m/rad. 0 increases in time according to 0 = ať²/2, where a is a constant.
(a) Sketch the motion, and indicate the approximate velocity and acceleration at three different points.
![Hints: a) The trajectory is given, so to sketch velocity and acceleration vectors at three different points
find the velocity and acceleration (for acceleration a see hints given in part b).
v = ř f + rê ð and a = (* – rở ) î + (rë + 2řė) ôð
The components of velocity and acceleration in polar coordinates should give a feeling about magnitude and direction of vectors
v and a as function of position. You do not need to be perfect in the sketch , just give e feeling (i.e., do not calculate) how the
magnitude and direction of these vectors change as function of position . Remember velocity is always in the tangential direction
with the trajectory, which in our case is a spiral one and not a circular one
from the formula of the acceleration in polar coordinates prove that
5a²f²
b)
a = (* – ri) î + (rë + 2řė) ôð = (¤
2л
2л
5a?f
) ê +
2л
() (1 – 20) F + 50 ê]
a=(
2л](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd8694f02-79fe-4b76-9079-b1456b2e1bd2%2F76f587e4-0183-4353-9c91-22d8e7ef04d7%2F9wvra4g_processed.png&w=3840&q=75)
Transcribed Image Text:Hints: a) The trajectory is given, so to sketch velocity and acceleration vectors at three different points
find the velocity and acceleration (for acceleration a see hints given in part b).
v = ř f + rê ð and a = (* – rở ) î + (rë + 2řė) ôð
The components of velocity and acceleration in polar coordinates should give a feeling about magnitude and direction of vectors
v and a as function of position. You do not need to be perfect in the sketch , just give e feeling (i.e., do not calculate) how the
magnitude and direction of these vectors change as function of position . Remember velocity is always in the tangential direction
with the trajectory, which in our case is a spiral one and not a circular one
from the formula of the acceleration in polar coordinates prove that
5a²f²
b)
a = (* – ri) î + (rë + 2řė) ôð = (¤
2л
2л
5a?f
) ê +
2л
() (1 – 20) F + 50 ê]
a=(
2л
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