In the mechanism shown, the angular velocity of link AB is 5 rad/s anticlockwise. At the instant shown, determine the angular velocity of link BC and the velocity of piston C. (asked in MU Dec 2025)Figure (described in words): Link AB hangs verticall…

Answer: \(\omega_{BC}=6.67\) rad/s (clockwise); \(v_C=3\) m/s, vertically downward.

Velocity of B: A is fixed and AB is vertical, so B moves perpendicular to AB, i.e. horizontally: \(v_B=\omega_{AB}\times AB=5\times0.2=1\) m/s. With AB anticlockwise and B below A, \(v_B\) is directed to the right.

Instantaneous centre (I) of BC:

• At B, draw a line perpendicular to \(v_B\): the vertical line through B (line AB extended downward).
• At C, the piston moves vertically, so draw a line perpendicular to \(v_C\): the horizontal line through C.
• These meet at I, directly below B at the level of C. So \(IB=150\) mm (vertical) and \(IC=450\) mm (horizontal).

Angular velocity of BC:

\[\omega_{BC}=\frac{v_B}{IB}=\frac{1}{0.15}=6.67\ \text{rad/s}\]

B is above I and moves to the right, so BC rotates clockwise about I.

Velocity of piston C:

\[v_C=\omega_{BC}\times IC=6.67\times0.45=3\ \text{m/s}\]

C is to the right of I and the rotation is clockwise, so C moves vertically downward.