Question

If rod has linear mass density $\lambda = \frac{m}{l}(1+\frac{x}{l})$, where x is the distance from point mass m. $V_a$ and $V_b$ are variable velocity. Find velocity of centre of mass.

Answer 1

$\frac{10}{27} V_a + \frac{17}{27} V_b$

Answer 2

The problem asks for the velocity of the center of mass of a system consisting of two point masses and a connecting rod with a variable linear mass density.

1. Calculate the mass of the rod ($M_{rod}$): The linear mass density is given by $\lambda(x) = \frac{m}{L}\left(1 + \frac{x}{L}\right)$, where $x$ is the distance from the point mass $m$ (let's call it point A) and $L$ is the length of the rod. The mass of an infinitesimal element $dx$ of the rod is $dm = \lambda(x) dx$. To find the total mass of the rod, we integrate $\lambda(x)$ from $x=0$ to $x=L$: $M_{rod} = \int_{0}^{L} \lambda(x) dx = \int_{0}^{L} \frac{m}{L}\left(1 + \frac{x}{L}\right) dx$ $M_{rod} = \frac{m}{L} \left[ x + \frac{x^2}{2L} \right]_{0}^{L}$ $M_{rod} = \frac{m}{L} \left( L + \frac{L^2}{2L} \right) = \frac{m}{L} \left( L + \frac{L}{2} \right) = \frac{m}{L} \left( \frac{3L}{2} \right)$ $M_{rod} = \frac{3m}{2}$

2. Calculate the total mass of the system ($M_{total}$): The system consists of point mass A ($m_A = m$), point mass B ($m_B = 2m$), and the rod ($M_{rod} = \frac{3m}{2}$). $M_{total} = m_A + m_B + M_{rod} = m + 2m + \frac{3m}{2} = 3m + \frac{3m}{2} = \frac{6m + 3m}{2} = \frac{9m}{2}$

3. Determine the velocity profile of the rod: Assuming the rod is rigid, the velocity of any point on the rod at a distance $x$ from mass $m$ can be expressed as a linear interpolation of the velocities of its ends. Let $\vec{V}_A$ be the velocity of mass $m$ and $\vec{V}_B$ be the velocity of mass $2m$. The velocity of an infinitesimal element $dm$ at position $x$ is: $\vec{v}(x) = \left(1 - \frac{x}{L}\right)\vec{V}_A + \frac{x}{L}\vec{V}_B$

4. Calculate the total momentum of the rod: The momentum of the rod is $\int \vec{v}(x) dm = \int_{0}^{L} \left[ \left(1 - \frac{x}{L}\right)\vec{V}_A + \frac{x}{L}\vec{V}_B \right] \lambda(x) dx$ Substitute $\lambda(x) = \frac{m}{L}\left(1 + \frac{x}{L}\right)$: $\int_{0}^{L} \left[ \left(1 - \frac{x}{L}\right)\vec{V}_A + \frac{x}{L}\vec{V}_B \right] \frac{m}{L}\left(1 + \frac{x}{L}\right) dx$ $= \frac{m}{L} \int_{0}^{L} \left[ \left(1 - \frac{x^2}{L^2}\right)\vec{V}_A + \left(\frac{x}{L} + \frac{x^2}{L^2}\right)\vec{V}_B \right] dx$ $= \frac{m}{L} \vec{V}_A \int_{0}^{L} \left(1 - \frac{x^2}{L^2}\right) dx + \frac{m}{L} \vec{V}_B \int_{0}^{L} \left(\frac{x}{L} + \frac{x^2}{L^2}\right) dx$

Evaluate the integrals: $\int_{0}^{L} \left(1 - \frac{x^2}{L^2}\right) dx = \left[ x - \frac{x^3}{3L^2} \right]_{0}^{L} = L - \frac{L^3}{3L^2} = L - \frac{L}{3} = \frac{2L}{3}$ $\int_{0}^{L} \left(\frac{x}{L} + \frac{x^2}{L^2}\right) dx = \left[ \frac{x^2}{2L} + \frac{x^3}{3L^2} \right]_{0}^{L} = \frac{L^2}{2L} + \frac{L^3}{3L^2} = \frac{L}{2} + \frac{L}{3} = \frac{3L + 2L}{6} = \frac{5L}{6}$

So, the total momentum of the rod is: $\frac{m}{L} \vec{V}_A \left( \frac{2L}{3} \right) + \frac{m}{L} \vec{V}_B \left( \frac{5L}{6} \right) = \frac{2m}{3} \vec{V}_A + \frac{5m}{6} \vec{V}_B$

5. Calculate the velocity of the center of mass ($V_{CM}$): The velocity of the center of mass for a system is given by: $\vec{V}_{CM} = \frac{\sum m_i \vec{V}_i + \int \vec{v}_{dm} dm}{M_{total}}$ $\vec{V}_{CM} = \frac{m_A \vec{V}_A + m_B \vec{V}_B + \left( \frac{2m}{3} \vec{V}_A + \frac{5m}{6} \vec{V}_B \right)}{M_{total}}$ $\vec{V}_{CM} = \frac{m \vec{V}_A + 2m \vec{V}_B + \frac{2m}{3} \vec{V}_A + \frac{5m}{6} \vec{V}_B}{\frac{9m}{2}}$

Combine the terms for $\vec{V}_A$ and $\vec{V}_B$: Coefficient of $\vec{V}_A$: $m + \frac{2m}{3} = \frac{3m + 2m}{3} = \frac{5m}{3}$ Coefficient of $\vec{V}_B$: $2m + \frac{5m}{6} = \frac{12m + 5m}{6} = \frac{17m}{6}$

So, the numerator is: $\frac{5m}{3} \vec{V}_A + \frac{17m}{6} \vec{V}_B$

Now, substitute this into the $V_{CM}$ formula: $\vec{V}_{CM} = \frac{\frac{5m}{3} \vec{V}_A + \frac{17m}{6} \vec{V}_B}{\frac{9m}{2}}$ $\vec{V}_{CM} = \frac{m \left( \frac{5}{3} \vec{V}_A + \frac{17}{6} \vec{V}_B \right)}{m \left( \frac{9}{2} \right)}$ $\vec{V}_{CM} = \frac{2}{9} \left( \frac{5}{3} \vec{V}_A + \frac{17}{6} \vec{V}_B \right)$ $\vec{V}_{CM} = \left( \frac{2}{9} \times \frac{5}{3} \right) \vec{V}_A + \left( \frac{2}{9} \times \frac{17}{6} \right) \vec{V}_B$ $\vec{V}_{CM} = \frac{10}{27} \vec{V}_A + \frac{17}{27} \vec{V}_B$

The final answer is $\frac{10}{27} V_a + \frac{17}{27} V_b$.

Explanation of the solution:

1. Calculate Rod Mass: Integrate the given linear mass density $\lambda(x)$ over the length of the rod to find its total mass $M_{rod}$.
2. Calculate Total System Mass: Sum the masses of the two point masses and the calculated mass of the rod to get $M_{total}$.
3. Determine Rod Velocity Profile: Assume the rod is rigid. The velocity of any point on a rigid rod connecting two points A and B, moving with velocities $\vec{V}_A$ and $\vec{V}_B$ respectively, is given by linear interpolation: $\vec{v}(x) = (1 - x/L)\vec{V}_A + (x/L)\vec{V}_B$.
4. Calculate Rod's Momentum Contribution: Integrate the product of the velocity profile $\vec{v}(x)$ and the differential mass $dm = \lambda(x)dx$ over the rod's length. This yields the total momentum of the rod.
5. Apply Center of Mass Velocity Formula: Use the general formula for the velocity of the center of mass of a system: $\vec{V}_{CM} = \frac{\sum m_i \vec{V}_i}{M_{total}}$. Sum the individual momenta of the point masses ($m_A \vec{V}_A$ and $m_B \vec{V}_B$) and the calculated total momentum of the rod, then divide by the total mass of the system.