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# Differentiation

Anonymous
wrote...
A month ago
 Differentiation 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 Attached file  Thumbnail(s): You must login or register to gain access to this attachment. Read 95 times 2 Replies
Replies
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.
AnonymousAnonymous
wrote...
3 weeks ago