Collar B moves to the left with a speed of 5 ms which is inc

Collar B moves to the left with a speed of 5 m/s, which is increasing at a constant rate of 1.5 m/s^2, relative to the hoop. The hoop rotates with the angular velocity and angular acceleration shown. Determine the magnitudes of the velocity and acceleration of the collar at this instant.

Solution

Let the rotating reference fram xyz be fixed to the hoop and let it conincide with the fixed reference frame XYZ a the instant considered.

Origin at A. X and x axis horizontal towards right. Y and y axes are vertically upwards.

The motion of the xyz frame with respect to the XYZ frame is

v_a = 0,

a_a = 0,

w = -6k rad/s

alpha = -3k rad/s^2

For the motion of collar B with respect to the xyz frame,

r_b/a = -0.45j m

(v_rel)xyz = -5i m/s

The normal components of (a_rel)xyz is

[(a_rel)xyz]_n = ((v_rel)xyz)^2 / r

= 5^2 / 0.2

= 125 m/s^2

Thus, (a_rel)xyz = -1.5i + 125j m/s

Now, applying the relative velocity equation,

v_b = v_a + w X r_b/a + (v_rel)xyz

= 0 + (-6k) X (-0.45j) + (-5i)

= -7.7i m/s

Thus v_b is 7.7 m/s towards left.

Applying the relative acceleration equation,

a_b = [a_a] + [alpha X r_b/a] + [w X (w X r_b/a)] + [2w X (v_rel)xyz] + [(a_rel)xyz]

= 0 + (-3k) X (-0.45j) + (-6k) X [(-6k) X (-0.45j)] + 2(-6k) X (-5i) + (-1.5i + 125j)

= -2.85i + 201.2j m/s^2

Thus, magnitude of a_b is therefore, sqrt(2.85^2 + 201.2^2) = 201 m/s^2