Question

Determine the magnitude of the resultant force acting on a 5 −kg particle at the instant t=2 s, if the particle is moving along a horizontal path defined by the equations r=(2t+10) m and θ=(1.5t2−6t) rad, where t is in seconds. Express your answer with the appropriate units.

Answer 1

Answer:

F = 296.7N

Explanation:

The x and y component of the position vector are given by:

$x(t) =rcos\phi, y(t) = rsin\phi\\ \frac{dx}{dt} =\frac{dr}{dt} cos\phi - rsin\phi\frac{d\phi}{dt} , \frac{dy}{dt}=\frac{dr}{dt}sin\phi + rcos\phi\frac{d\phi}{dt}\\ \frac{d^2x}{dt^2} = \frac{d^2r}{dt^2} cos\phi - \frac{dr}{dt} sin\phi\frac{d\phi}{dt} -\frac{dr}{dt} sin\phi\frac{d\phi}{dt} - r(cos\phi\frac{d^2\phi}{dt^2} + sin\phi\frac{d^2\phi}{dt^2})$

$\frac{d^2y}{dt^2} = \frac{d^2r}{dt^2} sin\phi + \frac{dr}{dt} cos\phi\frac{d\phi}{dt}+\frac{dr}{dt}cos\phi\frac{d\phi}{dt}+r(-sin\phi\frac{d^2\phi}{dt^2} +cos\phi\frac{d^2\phi}{dt^2})$

At t = 2s:

$\phi(2) = 1.5t^2-6t= -6\\ \frac{d\phi}{dt}(2) = 3t-6=0\\ \frac{d^2\phi}{dt^2}=3\\r(2)=2t+10=14\\ \frac{dr}{dt}=2\\\frac{d^2r}{dt^2} = 0$

Plugging in:

$\frac{d^2x}{dt^2}=-42(cos(-6) + sin(-6))=-52\frac{m}{s^2} \\\frac{d^2y}{dt^2} = 42(cos(-6)-sin(-6))=28.6\frac{m}{s^2}$

The resulting force F is:

$F = m\sqrt{(\frac{d^2x}{dt^2})^2 + (\frac{d^2y}{dt^2})^2}=296.7N$