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Anonymous joshephn
wrote...
A month ago
Full solution pls ?

The position of an object subject to uniform circular motion around the origin is given by (see
Vector Il worksheet, Week 4):
cos(0) t)
r(t)=R
sin(øt)
(a) Calculate the velocity and acceleration vectors for this trajectory.
(b) Show that a(t) is perpendicular to v(t) for all t.
(c) Show that a(t)• r(t) for all t. Hint: Recall that 1
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pie
wrote...
A month ago
(a) The position vector r(t) is given by r(t) = R cos(wt)i + R sin(wt)j.


To find the velocity vector v(t), we differentiate r(t) with respect to time t.


v(t) = dr(t)/dt = -Rw sin(wt)i + Rw cos(wt)j.


Similarly, to find the acceleration vector a(t), we differentiate v(t) with respect to time t.


a(t) = dv(t)/dt = -Rw^2 cos(wt)i - Rw^2 sin(wt)j.


(b) To show that a(t) is perpendicular to v(t) for all t, we take the dot product of a(t) and v(t). If the dot product is zero, then the vectors are perpendicular.


a(t) . v(t) = (-Rw^2 cos(wt)i - Rw^2 sin(wt)j) . (-Rw sin(wt)i + Rw cos(wt)j)
= Rw^3 cos(wt) sin(wt) - Rw^3 sin(wt) cos(wt)
= 0.


Since the dot product is zero, a(t) is perpendicular to v(t) for all t.


(c) To show that a(t) . r(t) = -w^2R^2 for all t, we take the dot product of a(t) and r(t).


a(t) . r(t) = (-Rw^2 cos(wt)i - Rw^2 sin(wt)j) . (R cos(wt)i + R sin(wt)j)
= -R^2w^2 cos^2(wt) - R^2w^2 sin^2(wt)
= -R^2w^2 (cos^2(wt) + sin^2(wt))
= -R^2w^2 * 1
= -w^2R^2.


Therefore, a(t) . r(t) = -w^2R^2 for all t.
Answer accepted by topic starter
AnonymousAnonymous
wrote...
3 weeks ago
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