Solution: Instantaneous Centre of Rotation
Problem: A rigid lamina is sliding on a horizontal floor (x-y plane). At an instant, the velocity components of three particles A, B, and C are given:
- Particle A at (0, 0): vAx = 2 m/s
- Particle B at (1, 2): vBx = 4 m/s
- Particle C at (2, 1): vCy = -2 m/s (Corrected)
1. Understanding the Concept
The Instantaneous Centre of Rotation (ICR), denoted as I(x0, y0), is a point about which the entire rigid body appears to be rotating at that instant with an angular velocity ω.
The velocity components of any point (x, y) on the body are related to the ICR by:
- vx = –ω(y – y0)
- vy = ω(x – x0)
2. Applying to Particle A (0, 0)
Given vAx = 2 m/s. Using the vx formula:
2 = –ω(0 – y0)
2 = ωy0
(Equation 1)
3. Applying to Particle B (1, 2)
Given vBx = 4 m/s. Using the vx formula:
4 = –ω(2 – y0)
4 = -2ω + ωy0
Substitute ωy0 = 2 from Equation 1:
4 = -2ω + 2
2ω = -2
ω = -1 rad/s
Now find y0 using Equation 1:
(-1)y0 = 2
y0 = -2
4. Applying to Particle C (2, 1) to find x0
Given vCy = -2 m/s. Using the vy formula:
-2 = ω(2 – x0)
Substitute ω = -1:
-2 = -1(2 – x0)
Dividing both sides by -1:
2 = 2 – x0
x0 = 0
Visualization of Motion
Fig 1: The green point is the ICR at (0,-2). The red arrows are velocity vectors.
Coordinates of ICR = (0, -2) m