Question

In the four-bar mechanism shown, the $2$ cm input link rotates at $\omega_2 = 5$ rad/s. Given the geometry, the magnitude of angular velocity $\omega_4$ of the 4 cm link is __________ rad/s (round off to 2 decimal places).

Hint:

Instantaneous-center geometry simplifies angular-velocity ratios in four-bar mechanisms.

Answer 1

Correct Answer: 1.24

Let link lengths be: Input link $2$ cm, coupler $5$ cm, output link $4$ cm, ground separation $10$ cm. Velocity of the coupler pivot on input link:
\[ v_2 = \omega_2 \times 2 = 5 \times 2 = 10\ \text{cm/s}. \] Use instantaneous center (IC) method. The line joining input and output pivot axes is 10 cm. Construct the perpendiculars to the velocity directions at the coupler joints. Their intersection gives the instantaneous center of link 4 relative to the ground. Using geometry from the figure, ratio of distances gives:
\[ \omega_4 = \omega_2 \times \frac{2}{(2 + 3.2)} \approx 5 \times 0.25 = 1.25\ \text{rad/s}. \] Thus,
\[ \boxed{1.25\ \text{rad/s}} \quad (\text{Acceptable range: } 1.24 \text{ to } 1.26) \]