Title: Differentiation Post by: joshephn on Apr 26, 2024 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 Title: Re: Differentiation Post by: pie on Apr 26, 2024 (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. Title: Re: Differentiation Post by: bio_man on May 1, 2024 Content hidden
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