Find the instantaneous axis of rotation of a rod of length l when its end A moves with velocity v_A = v î and the rod rotates with angular velocity ω = - (v / 2l) k̂.

To find the instantaneous axis of rotation of a rotating rod, we need to consider both the linear velocity of the point AA and the angular velocity of the rod.
Given:
• The velocity of the point AA is vA=vi^\mathbf{v}_A = v \hat{i}.
• The angular velocity of the rod is ω=−v2lk^\boldsymbol{\omega} = -\frac{v}{2l} \hat{k}.
• The length of the rod is ll.
Step 1: Relationship between velocity and angular velocity
For a rotating body, the velocity of any point on the body is related to the angular velocity by the following equation:
v=ω×r\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}
Where:
• v\mathbf{v} is the velocity of the point,
• ω\boldsymbol{\omega} is the angular velocity vector,
• r\mathbf{r} is the position vector of the point relative to the axis of rotation.
For the point AA, we have:
• The position of AA relative to the origin (since the rod's one end is at AA and the other end is at BB) is rA=li^\mathbf{r}_A = l \hat{i} (assuming AA is at the origin of the coordinate system).
The velocity at point AA is given as vA=vi^\mathbf{v}_A = v \hat{i}.
Step 2: Finding the instantaneous axis of rotation
The instantaneous axis of rotation is the point at which the velocity is zero. This means that the position of the instantaneous axis is such that the velocity of every point on the rod is perpendicular to the position vector with respect to that axis.
We can find the location of the instantaneous axis of rotation using the relationship between linear velocity and angular velocity. Since the rod rotates with angular velocity ω=−v2lk^\boldsymbol{\omega} = -\frac{v}{2l} \hat{k}, this means the rod is rotating about a point that lies perpendicular to the angular velocity vector.
Let’s use the following formula to find the position of the instantaneous center of rotation (ICR), or the point where the velocity is zero:
ICR=vA×ω∣ω∣2\text{ICR} = \frac{\mathbf{v}_A \times \boldsymbol{\omega}}{|\boldsymbol{\omega}|^2}
Step 3: Calculate the cross product
Now, let’s compute the cross product vA×ω\mathbf{v}_A \times \boldsymbol{\omega}:
• vA=vi^\mathbf{v}_A = v \hat{i},
• ω=−v2lk^\boldsymbol{\omega} = -\frac{v}{2l} \hat{k}.
The cross product is:
vA×ω=vi^×(−v2lk^)\mathbf{v}_A \times \boldsymbol{\omega} = v \hat{i} \times \left(-\frac{v}{2l} \hat{k}\right)
Using the right-hand rule, the cross product of i^\hat{i} and k^\hat{k} gives j^\hat{j}, so we have:
vA×ω=−v22lj^\mathbf{v}_A \times \boldsymbol{\omega} = -\frac{v^2}{2l} \hat{j}
Step 4: Calculate the magnitude of ω\boldsymbol{\omega}
The magnitude of the angular velocity is:
∣ω∣=∣v2l∣=v2l|\boldsymbol{\omega}| = \left|\frac{v}{2l}\right| = \frac{v}{2l}
Now, ∣ω∣2|\boldsymbol{\omega}|^2 is:
∣ω∣2=(v2l)2=v24l2|\boldsymbol{\omega}|^2 = \left(\frac{v}{2l}\right)^2 = \frac{v^2}{4l^2}
Step 5: Find the position of the instantaneous center of rotation (ICR)
Now, use the formula to find the position of the instantaneous axis of rotation (ICR):
ICR=vA×ω∣ω∣2\text{ICR} = \frac{\mathbf{v}_A \times \boldsymbol{\omega}}{|\boldsymbol{\omega}|^2}
Substitute the values:
ICR=−v22lj^v24l2\text{ICR} = \frac{-\frac{v^2}{2l} \hat{j}}{\frac{v^2}{4l^2}}
Simplifying:
ICR=−2l1j^=−2lj^\text{ICR} = -\frac{2l}{1} \hat{j} = -2l \hat{j}
This means that the instantaneous axis of rotation is located a distance of 2l2l along the negative yy-axis.
Final Answer:
The instantaneous axis of rotation of the rod lies along the yy-axis at a distance of 2l2l below the point AA. Thus, the instantaneous axis is at the position −2lj^-2l \hat{j}.