In one line
Use the crank's known rotation to get the velocity of the crank pin, then locate the connecting rod's ICR from the crank-pin and slider velocity directions.
Explained simply
In a car engine or a hand pump, a rotating crank pushes a rod, which pushes a piston back and forth. Knowing how fast the crank turns, we want the piston speed at a given position. The ICR turns this into simple geometry.
Technically, the crank pin A has velocity \( v_A = \omega_{OA} \cdot OA \), perpendicular to OA. The slider B moves along its guide. The rod AB's ICR is where the line OA (extended) meets the perpendicular to the guide at B.
How it works
- Draw the mechanism at the given crank angle; find all lengths and angles (sine rule).
- \( v_A = \omega_{OA} \times OA \), perpendicular to OA.
- Perpendicular to \( v_A \) is the line OA itself; extend it.
- Perpendicular to \( v_B \) (along the guide) is a line at B normal to the guide.
- Intersection = I. Measure IA and IB.
- \( \omega_{AB} = v_A / IA \); \( v_B = \omega_{AB} \cdot IB \).
- If the crank is perpendicular to the guide, the two perpendiculars are parallel, I is at infinity, \( \omega_{AB} = 0 \) and \( v_B = v_A \).
Key formula
\( v_A = \omega_{OA} \cdot OA \); \( \omega_{AB} = v_A/IA \); \( v_B = \omega_{AB} \cdot IB \). Lengths in m, \( \omega \) in rad/s.
Quick example
Crank OA = 0.2 m at 10 rad/s, rod AB = 0.6 m, crank at 45°. A is at (0.1414, 0.1414) m; B is at x = 0.1414 + \( \sqrt{0.36 - 0.02} \) = 0.7245 m. I lies on line OA at x = 0.7245, so I = (0.7245, 0.7245). IA = 0.8246 m, IB = 0.7245 m. \( v_A = 2 \) m/s, \( \omega_{AB} = 2/0.8246 = 2.43 \) rad/s, \( v_B = 2.43 \times 0.7245 = 1.76 \) m/s.
Common mistakes
- Drawing the perpendicular to \( v_A \) at A but not through O; for a crank it is the line OA itself.
- Using \( \omega \) of the crank for the rod; each link has its own \( \omega \).
- Forgetting the special 'ICR at infinity' case.
How MU asks it
6 to 8-mark numerical: 'In the mechanism shown, crank OA rotates at N rpm; find the angular velocity of AB and the velocity of slider B'. Convert rpm, draw the figure, locate I, show the sine-rule geometry and give results with directions. Mechanisms with more than 2 links are outside the syllabus.