The position of point A on the crane is described by polar coordinates (r, theta

Question

The position of point A on the crane is described by polar coordinates (r, theta). the distance of point AS from the origin is prescribed by a given function, that is r = f(theta (f)). Describe expression for the position, velocity, and acceleration vectors of:theta and/or any of its time derivatives the function f and/or any of its derivatives with respect to theta, and the unit vectors e, e_theta. If point A starts at theta -= 0 and the angular velocity is given by theta = 0.05 t rad/sec, and f = 5 + 2 theta^2 9in meters), determine the velocity and acceleration vectors when theta = pi/4.

Explanation / Answer

r=f(theta(t))

position vector=f(theta(t)) er + theta(t) e_theta

where er and e_theta are unit vectors

velocity vector=time derivative of position vector

=f'(theta(t))*theta'(t) er+ theta'(t) e_theta

where theta'(t) is first derivative of theta(t)

acceleration vector=time derivative of velocity vector

=(f''(theta(t))*theta'(t)+f'(theta(t))*(theta'(t))^2) er+theta''(t) e_theta

part b:

velocity vector

=f'(theta(t))*theta'(t) er+ theta'(t) e_theta
=4*theta*0.05*t er+ 0.05*t e_theta

now given theta'(t)=0.05*t

==>theta(t)=0.05*t*dt=0.025*t^2+c

as at t=0, theta=0, ==>c=0

hence theta(t)=0.025*t^2

so when theta=pi/4

t=5.605 seconds

so velocity vector=4*(pi/4)*0.05*5.605 er+ 0.05*5.605 e_theta

=0.88043 er+ 0.28025 e_theta

acceleration vector=(4*0.05*t+4*theta*(0.05*t)^2) er+ 0.05 e_theta

at t=5.605 seconds,

acceleration vector=(4*0.05*5.605+4*(pi/4)*(0.05*5.605)^2) er+ 0.05 e_theta

=1.3677 er+ 0.05 e_theta

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