Question

Rod A of mass m, length $l$ and coefficient of thermal coefficient $\alpha$ is joined with Rod B of mass 2m, length $l$ and coefficient of thermal expansion $\frac{\alpha}{3}$. Rods are kept on smooth floor. If temperature of rods is increased by $\Delta T$, then junction of the rods shifts by distance $\frac{\alpha l \Delta T}{n}$, Find value of n.

Answer 1

18

Answer 2

Solution:

Let the initial positions be such that rod A extends from 0 to $l$ (with mass $m$) and rod B from $l$ to $2l$ (with mass $2m$). When heated by $\Delta T$:

1. Expansion of lengths:

   • Rod A expands from $l$ to $$l_A = l(1+\alpha\Delta T).$$
   • Rod B expands from $l$ to $$l_B = l\left(1+\frac{\alpha}{3}\Delta T\right).$$

2. Displacements:

   Let $u$ be the displacement of the free (left) end of rod A, $s$ the displacement of the junction (which is common to both rods) and $v$ the displacement of the free (right) end of rod B. Then

   $$s - u = l + \alpha l\Delta T \quad \text{(rod A)}$$

   and

   $$v - s = l + \frac{\alpha l\Delta T}{3} \quad \text{(rod B)}.$$

3. Conservation of center of mass (COM):

   Since the floor is smooth, no external horizontal force acts; hence the overall COM remains fixed. The new COM positions of the rods (being uniform) are the averages of their endpoints:

   $$\text{COM}_A = \frac{u+s}{2},\quad \text{COM}_B = \frac{s+v}{2}.$$

   The overall COM of the system (total mass $3m$) must equal its initial value:

   $$\frac{m\frac{(u+s)}{2} + 2m\frac{(s+v)}{2}}{3m} = \text{constant}.$$

   Initially, COM of rod A is $\frac{l}{2}$ and rod B is $\frac{3l}{2}$, so overall COM is:

   $$\frac{m\frac{l}{2} + 2m\frac{3l}{2}}{3m} = \frac{\frac{l}{2}+3l}{3} = \frac{7l}{6}.$$

4. Expressing endpoints in terms of $s$:

   From the expansion conditions:

   $$u = s - \left(l+\alpha l\Delta T\right),$$ $$v = s + \left(l+\frac{\alpha l\Delta T}{3}\right).$$

5. Setting up COM conservation:

   The new overall COM is

   $$\frac{1}{6}\left[(u+s) + 2(s+v)\right] = \frac{1}{6}\Bigl[ \bigl(s - (l+\alpha l\Delta T) + s\bigr) + 2\Bigl(s+s + l+\frac{\alpha l\Delta T}{3}\Bigr)\Bigr].$$

   Simplify step‐by‐step:

   $$u+s = 2s - \left(l+\alpha l\Delta T\right),$$ $$2(s+v) = 2\left(2s + l+\frac{\alpha l\Delta T}{3}\right) = 4s + 2l + \frac{2\alpha l\Delta T}{3}.$$

   Therefore,

   $$\text{New COM} = \frac{6s + l - \frac{\alpha l\Delta T}{3}}{6} = s + \frac{l}{6} - \frac{\alpha l\Delta T}{18}.$$

   Equating to the initial COM:

   $$s + \frac{l}{6} - \frac{\alpha l\Delta T}{18} = \frac{7l}{6}.$$

   Solve for $s$:

   $$s = \frac{7l}{6} - \frac{l}{6} +\frac{\alpha l\Delta T}{18} = l + \frac{\alpha l\Delta T}{18}.$$

6. Junction shift:

   The junction originally was at $x=l$ and now is at

   $$s = l + \frac{\alpha l\Delta T}{18}.$$

   So the shift is

   $$\Delta x = \frac{\alpha l\Delta T}{18}.$$

   Comparing with the given form $\frac{\alpha l\Delta T}{n}$, we identify

   $$n = 18.$$

Explanation (minimal):

• Write expansion conditions: $s-u = l+\alpha l\Delta T$ and $v-s = l+\frac{\alpha l\Delta T}{3}$.
• Express endpoints $u$ and $v$ in terms of $s$.
• Use COM conservation: $$\frac{(u+s) + 2(s+v)}{6} = \frac{7l}{6}.$$
• Solve to obtain $s = l + \frac{\alpha l\Delta T}{18}$.
• Thus, junction shift $=\frac{\alpha l\Delta T}{18}$ implying $n=18$.