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)| = √(\left(4b\omega \sin(\omega t)\right)^2+\left(3b\omega \cos(\omega t)\right)^2) \\ \\ = √(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=√(16b^2\omega^2\sin^2(\omega (0))+9b^2\omega^2\cos^2(\omega (0))) \\ \\ =√(16b^2\omega^2\sin^2(0)+9b^2\omega^2\cos^2(0))=√(16b^2\omega^2(0)+9b^2\omega^2(1)) \\ \\ = √(9b^2\omega^2) =\bold{3b\omega}`.

d) The speed v at
`t = (\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 ( (\pi)/(2\omega) ))+9b^2\omega^2\cos^2(\omega ( (\pi)/(2\omega) ))} \\ \\ =\sqrt{16b^2\omega^2\sin^2( (\pi)/(2) )+9b^2\omega^2\cos^2( (\pi)/(2) )}=√(16b^2\omega^2(1)+9b^2\omega^2(0)) \\ \\ = √(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)| = √(\left(4b\omega^2 \cos(\omega t)\right)^2+\left(3b\omega^2 \sin(\omega t)\right)^2) \\ \\ = √(16b^2\omega^4\cos^2(\omega t)+9b^2\omega^4\sin(\omega t))`.