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Relative Velocity and Acceleration withTranslating and Rotating Frames

Example: Block R slides to the left at a constant speed v_R while a bar B leans against it always

remaining in contact with the bar. Let P be the point in the bar in contact with point S, a point

in the block. The lower end of the bar, Q, has a velocity v_Q i m/sec and an acceleration a_Q i m/sec².

Let xy be a fixed frame of reference with unit vectors i and j . Let x₁y₁ be a translating/rotating

frame with origin at Q attached to the bar with unit vectors i₁ and j₁ . See the figure below.

The following data apply: d = 4 m, h = 3 m, v_R = - 5 m/sec, , a_R = 0, v_Q = 20 m/sec, a_Q = - 36 m/sec²

Find the acceleration of P in the fixed frame of reference F. i.e. a_P|_F

Find the acceleration of P relative to the bar in terms of components in the fixed frame. i.e. a_P|_B

Find the angular acceleration of the bar.

Since P is a point in the bar coincident with S, a point in the block and block R is sliding ,

the acceleration of P is a_R i but a_R = 0 so a_P|_F = 0 i m/sec² (result)

To find the acceleration of P relative to the bar and the angular acceleration of the bar you need to

apply the general relative acceleration equation for a translating/rotating frame of reference.

a_B = a_A + a_B|₁ + α x r_AB - ω² r_AB+ 2 ω x v_B|₁

Relative Acceleration Equation

For this example the equation becomes (written for points P and Q )

a_P|_F = a_Q|_F + a_P|_B+ α_B x r_QP - ω_B² r_QP+ 2 ω_B x v_P|_B

Note: The values of ω_B and of v_P|_B come from the relative velocity analysis for this

example. The results were: ω_B = 3 rad/sec and v_P|_B = ˗ 20 i₁ m/sec