Question

The figure shows a wheel rolling without slipping on a horizontal plane with angular velocity \(\omega_1\). A rigid bar PQ is pinned to the wheel at P while the end Q slides on the floor. What is the angular velocity \(\omega_2\) of the bar PQ? 

• A. \(\omega_2 = 2\omega_1\)
• B. \(\omega_2 = \omega_1\)
• C. \(\omega_2 = 0.5\omega_1\)
• D. \(\omega_2 = 0.25\omega_1\)

Hint:

The Instantaneous Center of Rotation (ICR) method can be faster here. The velocity of P is perpendicular to the line from R (ICR of the wheel) to P. The velocity of Q is horizontal. The ICR for the bar PQ is the intersection of the lines perpendicular to these velocities. The line perpendicular to \(\mathbf{v}_Q\) is a vertical line through Q. The line perpendicular to \(\mathbf{v}_P\) is the line RP itself. The ICR of PQ is the intersection of the vertical line through Q and the line RP. Once you find the location of this ICR, you can state \(v_P = \omega_2 \cdot (\text{dist from ICR to P})\) and solve for \(\omega_2\). However, the vector method is more robust if the geometry is complex.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a kinematics problem that can be solved using the concept of the Instantaneous Center of Rotation (ICR). For a body in general plane motion, the ICR is a point about which the body has pure rotational motion at that instant. The velocity of any point on the body is perpendicular to the line connecting the point to the ICR and is proportional to the distance from the ICR. Alternatively, we can use the relative velocity equation.
Step 2: Key Formula or Approach (Using Relative Velocity):
1. Find the absolute velocity of point P, \(\mathbf{v}_P\). Since P is on the rolling wheel, its velocity is the sum of the velocity of the wheel's center O and the velocity of P relative to O.
2. Find the absolute velocity of point Q, \(\mathbf{v}_Q\). Since Q slides on the floor, its velocity must be purely horizontal.
3. Use the relative velocity equation for the rigid bar PQ: \(\mathbf{v}_Q = \mathbf{v}_P + \mathbf{v}_{Q/P}\), where \(\mathbf{v}_{Q/P}\) is the velocity of Q relative to P, which is due to the rotation of bar PQ about P.
Step 3: Detailed Calculation:
Let's set up a coordinate system with \(\mathbf{\hat{i}}\) horizontal to the right and \(\mathbf{\hat{j}}\) vertically up. The origin can be at point R (the contact point of the wheel). The wheel rotates clockwise, so \(\boldsymbol{\omega}_1 = -\omega_1 \mathbf{\hat{k}}\). The bar PQ rotates counter-clockwise (from inspection), so \(\boldsymbol{\omega}_2 = \omega_2 \mathbf{\hat{k}}\).
1. Velocity of P (\(\mathbf{v}_P\)): - Velocity of center O: The wheel is rolling without slipping, so \(v_O = \omega_1 R\), where \(R\) is the wheel's radius. From the diagram, \(R=3\) m. So \(\mathbf{v}_O = (3\omega_1) \mathbf{\hat{i}}\).
- Position vector of P relative to O, \(\mathbf{r}_{P/O}\): From the diagram, P is at a distance of 2 m horizontally from the center. \(\mathbf{r}_{P/O} = 2 \mathbf{\hat{i}}\).
- Velocity of P relative to O: \(\mathbf{v}_{P/O} = \boldsymbol{\omega}_1 \times \mathbf{r}_{P/O} = (-\omega_1 \mathbf{\hat{k}}) \times (2 \mathbf{\hat{i}}) = -2\omega_1 (\mathbf{\hat{k}} \times \mathbf{\hat{i}}) = -2\omega_1 \mathbf{\hat{j}}\).
- Absolute velocity of P: \(\mathbf{v}_P = \mathbf{v}_O + \mathbf{v}_{P/O} = 3\omega_1 \mathbf{\hat{i}} - 2\omega_1 \mathbf{\hat{j}}\).
2. Velocity of Q (\(\mathbf{v}_Q\)):
- Q slides on the horizontal floor, so its velocity is purely horizontal: \(\mathbf{v}_Q = v_Q \mathbf{\hat{i}}\).
3. Relative Velocity Equation (\(\mathbf{v}_Q = \mathbf{v}_P + \mathbf{v}_{Q/P}\)): - Position vector of Q relative to P, \(\mathbf{r}_{Q/P}\): From the diagram, Q is 8 m to the right and 3 m below P's vertical level (since P is at height 3m and Q is at height 0). This is incorrect. P is at a distance of 2m from the center, so its height is 3m. Let's re-read the diagram. O is the center. Radius is 3m. P is on a horizontal line through O. So P is at (2, 3) relative to R. Q is at (10, 0) relative to R. Thus \(\mathbf{r}_{Q/P} = (10-2)\mathbf{\hat{i}} + (0-3)\mathbf{\hat{j}} = 8\mathbf{\hat{i}} - 3\mathbf{\hat{j}}\).
- \(\mathbf{v}_{Q/P} = \boldsymbol{\omega}_2 \times \mathbf{r}_{Q/P} = (\omega_2 \mathbf{\hat{k}}) \times (8\mathbf{\hat{i}} - 3\mathbf{\hat{j}}) = 8\omega_2(\mathbf{\hat{k}} \times \mathbf{\hat{i}}) - 3\omega_2(\mathbf{\hat{k}} \times \mathbf{\hat{j}}) = 8\omega_2 \mathbf{\hat{j}} - 3\omega_2 (-\mathbf{\hat{i}}) = 3\omega_2 \mathbf{\hat{i}} + 8\omega_2 \mathbf{\hat{j}}\).
4. Substitute and solve:
\[ v_Q \mathbf{\hat{i}} = (3\omega_1 \mathbf{\hat{i}} - 2\omega_1 \mathbf{\hat{j}}) + (3\omega_2 \mathbf{\hat{i}} + 8\omega_2 \mathbf{\hat{j}}) \] Group the \(\mathbf{\hat{i}}\) and \(\mathbf{\hat{j}}\) components: \[ v_Q \mathbf{\hat{i}} = (3\omega_1 + 3\omega_2) \mathbf{\hat{i}} + (-2\omega_1 + 8\omega_2) \mathbf{\hat{j}} \] For this vector equation to hold, the \(\mathbf{\hat{j}}\) component on the right side must be zero. \[ -2\omega_1 + 8\omega_2 = 0 \] \[ 8\omega_2 = 2\omega_1 \] \[ \omega_2 = \frac{2}{8}\omega_1 = \frac{1}{4}\omega_1 = 0.25\omega_1 \] Step 4: Final Answer:
The angular velocity of the bar PQ is \(\omega_2 = 0.25\omega_1\).
Step 5: Why This is Correct:
The relative velocity method provides a systematic way to relate the velocities of different points on a rigid body. By equating the components of the vector equation \(\mathbf{v}_Q = \mathbf{v}_P + \boldsymbol{\omega}_2 \times \mathbf{r}_{Q/P}\), we obtain a constraint on the angular velocities. The vertical component of \(\mathbf{v}_Q\) must be zero, which directly yields the relationship between \(\omega_1\) and \(\omega_2\).