A bar of length L = 2.5 m, is falling such that at the instant when ? = 34o
, vA = 3 m/s, and
aA = 8.7 m/s2
. At this instant, determine the angular velocity and angular acceleration of the bar
AB, as well as the acceleration of point D, which is the midpoint of the bar. A bar of length L =
2.5 m, is falling such that at the instant when ? = 34o , vA = 3 m/s, and aA = 8.7 m/s2 . At this
instant, determine the angular velocity and angular acceleration of the bar AB, as well as the
acceleration of point D, which is the midpoint of the bar.
Solution
taking downward direction as negative and anticlockwise rotation as positive
let point A be at y distance from the origin,
? y = l cos?
differentiating both sides wrt \'t\'
? (dy/dt) = l x (-sin?) x d?/dt
at ? = 34 degree, dy/dt = Va = -3m/s
? d?/dt = 2.145 rad/s = ?
differentiating again,
? (d2y/dt2) = -lcos? x d?/dt - lsin? x d2?/dt2
at ? = 34degree,
d2y/dt2 = aA= -8.7m/s2 , d2?/dt2 = ? ,d?/dt = 2.145rad/s
? -8.7 = - 2.5cos(34) x 2.145 -2.5sin(34) x ?
? ? = 3.043 rad/s2 ..........angular acceleration of the bar AB( anticlockwise).
by relative accleration relation.......
aD = aA + ? x rD/A
= -8.7 + 3.043 x 1.25
aD = -4.896 m/s2 ...................acceleration of point D (downward direction)..
Seal of Good Local Governance (SGLG) 2024Final.pptx
A bar of length L = 2.5 m, is falling such that at the instant when .pdf
1. A bar of length L = 2.5 m, is falling such that at the instant when ? = 34o
, vA = 3 m/s, and
aA = 8.7 m/s2
. At this instant, determine the angular velocity and angular acceleration of the bar
AB, as well as the acceleration of point D, which is the midpoint of the bar. A bar of length L =
2.5 m, is falling such that at the instant when ? = 34o , vA = 3 m/s, and aA = 8.7 m/s2 . At this
instant, determine the angular velocity and angular acceleration of the bar AB, as well as the
acceleration of point D, which is the midpoint of the bar.
Solution
taking downward direction as negative and anticlockwise rotation as positive
let point A be at y distance from the origin,
? y = l cos?
differentiating both sides wrt 't'
? (dy/dt) = l x (-sin?) x d?/dt
at ? = 34 degree, dy/dt = Va = -3m/s
? d?/dt = 2.145 rad/s = ?
differentiating again,
? (d2y/dt2) = -lcos? x d?/dt - lsin? x d2?/dt2
at ? = 34degree,
d2y/dt2 = aA= -8.7m/s2 , d2?/dt2 = ? ,d?/dt = 2.145rad/s
? -8.7 = - 2.5cos(34) x 2.145 -2.5sin(34) x ?
? ? = 3.043 rad/s2 ..........angular acceleration of the bar AB( anticlockwise).
by relative accleration relation.......
aD = aA + ? x rD/A
= -8.7 + 3.043 x 1.25
aD = -4.896 m/s2 ...................acceleration of point D (downward direction).
