PROPERTIES OF MATTER CYU 1

Solution 1

Problem 1: Thermal Creep on an Inclined Plane

θ During Heating (Expansion) Neutral Point f_upper (Down) f_lower (Up) mg sinθ

(a) Analysis of Temperature Rise (Expansion)

When the temperature increases by $\Delta T$, the rod expands. Friction opposes relative motion. There exists a “neutral point” on the rod where the displacement is zero relative to the plane.

  • Above Neutral Point ($l_1$): Rod expands upwards. Friction acts downwards.
  • Below Neutral Point ($l_2$): Rod expands downwards. Friction acts upwards.

The rod is in equilibrium, so the net friction force must balance the component of gravity acting down the slope ($mg \sin\theta$).

$$ f_{net} = f_{lower} – f_{upper} = mg \sin\theta $$

Assuming limiting friction density $f = \mu \lambda g \cos\theta$ (where $\lambda = m/L$):

$$ (\mu l_2 \lambda g \cos\theta) – (\mu l_1 \lambda g \cos\theta) = (\lambda L) g \sin\theta $$ $$ \mu (l_2 – l_1) \cos\theta = L \sin\theta \implies l_2 – l_1 = L \frac{\tan\theta}{\mu} $$

Solving with $l_1 + l_2 = L$, we find the lengths of the segments:

$$ l_1 = \frac{L}{2} \left( 1 – \frac{\tan\theta}{\mu} \right), \quad l_2 = \frac{L}{2} \left( 1 + \frac{\tan\theta}{\mu} \right) $$

Displacements:

  • Upper Edge: Shifts Up by $\Delta l_1 = l_1 \alpha \Delta T$.
  • Lower Edge: Shifts Down by $\Delta l_2 = l_2 \alpha \Delta T$.
  • Mass Center (COM): The COM is initially at $L/2$ from the top. The neutral point is at $l_1$. Since $l_1 < L/2$, the COM is located in the lower segment (moving down).
    Shift of COM = Distance from Neutral Point $\times \alpha \Delta T$
    Shift = $(L/2 – l_1) \alpha \Delta T = \frac{L}{2} \frac{\tan\theta}{\mu} \alpha \Delta T$ (Downwards).

(b) Analysis of Temperature Fall (Contraction)

During cooling, the rod contracts towards a new neutral point.

  • Upper part ($l_1’$): Moves down. Friction acts upwards.
  • Lower part ($l_2’$): Moves up. Friction acts downwards.
Net force equilibrium ($F_{net}$ balances $mg \sin\theta$): $$ F_{up} – F_{down} = mg \sin\theta $$ $$ \mu (l_1′ – l_2′) \cos\theta = L \sin\theta \implies l_1′ – l_2′ = L \frac{\tan\theta}{\mu} $$

Solving with $l_1′ + l_2′ = L$:

$$ l_1′ = \frac{L}{2} \left( 1 + \frac{\tan\theta}{\mu} \right) $$

Displacements:

  • Upper Edge: Shifts Down.
  • Lower Edge: Shifts Up.
  • Mass Center (COM): The neutral point is now lower down ($l_1′ > L/2$). The COM is above the neutral point, so it moves downwards.
    Shift = $(l_1′ – L/2) \alpha \Delta T = \frac{L \tan\theta}{2\mu} \alpha \Delta T$ (Downwards).

(c) Net Shift during Complete Cycle

The rod shifts downwards during heating and shifts further downwards during cooling. This phenomenon is known as “thermal creep”.

$$ \text{Total Shift} = \text{Shift}_{heating} + \text{Shift}_{cooling} $$ $$ \text{Total Shift} = \frac{L \tan\theta}{2\mu} \alpha \Delta T + \frac{L \tan\theta}{2\mu} \alpha \Delta T $$
Net Shift: $\frac{L \tan\theta}{\mu} \alpha \Delta T$ downwards along the incline.
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