Question

A small ball is fastened to a long rubber band and twirled around it such a way that the ball moves in an elliptical path given by the equation r(t) = i 4b cos(wt) + j 3b sin(wt), where b and w are positive constants. (

Answer 1

Given that a small ball is fastened to a long rubber band and twirled around it such a way that the ball moves in an elliptical path given by the equation
$r(t) = i\, 4b \cos(\omega t) + j\, 3b \sin(\omega t)$,
where b and $w$ are positive constants.

a) The velocity of the ball v as a function of time t is given by
$v=r'(t) = -i\, 4b\omega \sin(\omega t) + j\, 3b\omega \cos(\omega t)$.

b) The speed of the ball v = |v| as a function of t is given by
$|v|=|r'(t)| = \sqrt{\left(4b\omega \sin(\omega t)\right)^2+\left(3b\omega \cos(\omega t)\right)^2} \\ \\ = \sqrt{16b^2\omega^2\sin^2(\omega t)+9b^2\omega^2\cos^2(\omega t)}$.

c) The speed v at t = 0 at which time the ball is at its maximum distance from the origin is given by
$v=\sqrt{16b^2\omega^2\sin^2(\omega (0))+9b^2\omega^2\cos^2(\omega (0))} \\ \\ =\sqrt{16b^2\omega^2\sin^2(0)+9b^2\omega^2\cos^2(0)}=\sqrt{16b^2\omega^2(0)+9b^2\omega^2(1)} \\ \\ = \sqrt{9b^2\omega^2} =\bold{3b\omega}$.

d) The speed v at $t = \frac{\pi}{2\omega}$ at which time the ball is at its minimum distance from the origin is given by
$v=\sqrt{16b^2\omega^2\sin^2(\omega ( \frac{\pi}{2\omega} ))+9b^2\omega^2\cos^2(\omega ( \frac{\pi}{2\omega} ))} \\ \\ =\sqrt{16b^2\omega^2\sin^2( \frac{\pi}{2} )+9b^2\omega^2\cos^2( \frac{\pi}{2} )}=\sqrt{16b^2\omega^2(1)+9b^2\omega^2(0)} \\ \\ = \sqrt{16b^2\omega^2} =\bold{4b\omega}$.

e) The acceleration of the ball, a, as a function of t is given by
$a=r''(t) = -i\, 4b\omega^2 \cos(\omega t) - j\, 3b\omega^2 \sin(\omega t)$.

f) The magnitude of the acceleration of the ball
$|a|=|r''(t)| = \sqrt{\left(4b\omega^2 \cos(\omega t)\right)^2+\left(3b\omega^2 \sin(\omega t)\right)^2} \\ \\ = \sqrt{16b^2\omega^4\cos^2(\omega t)+9b^2\omega^4\sin(\omega t)}$.