Question

A circular disc made of iron is rotated about its axis at a constant velocity ω. Calculate the percentage change in the linear speed of a particle of the rim as the disc is slowly heated from 20°C to 50°C, keeping the angular velocity constant. Coefficient of linear expansion of iron = 1.2 × 10–5 °C–1.

Answer 1

The percentage change in the linear speed of a particle of the rim as the disc is slowly heated from 20°C to 50°C, is $\bold{3.6 \times 10^{-2}}$

Explanation:

Step 1:

$\alpha=1.2 \times 10^{-2} \mathrm{C}^{-1}$

Change in temperature

∆T=  30°C

Let R ' and R be the 50 C and 20 C particle radius, respectively.

If v and v ' are the particle's linear velocity at 50°C and 20°C, as the angular velocity remains (ω) constant.

$\omega=\mathrm{v} \mathrm{R}=\mathrm{v}^{\prime} \mathrm{R}^{\prime}$.....................(1)

Step 2:

we know that

$\mathrm{R}^{\prime}=\mathrm{R}(1+\alpha \Delta \mathrm{T})$

$\mathrm{R}^{\prime}=\mathrm{R}(1+\alpha \Delta \mathrm{T})$

$R^{\prime}=1.00036 R$

with equation(1) we have,

$\frac{v}{R}=\frac{v^{\prime}}{R^{\prime}}=\frac{v^{\prime}}{1.00036 R}$

${v}^{\prime}=1.00036 v$

Step 3:

The percentage change in linear speed is,

$=\left(\frac{1.00036 v-v}{v}\right) \times 100$

$=0.00036 \times 100$

$=3.6 \times 10^{-2}$

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

$3.6 \times {10}^{ - 2}$