Question

A uniform brass disc of radius $a$ and mass $m$ is set into spinning with angular speed ${\omega }_0$ about an axis passing through centre of disc and perpendicular to the plane of disc. If its temperature increases from ${{\theta }_1}^{\circ }C$ to ${{\theta }_2}^{\circ }C$ without distributing the disc, what will be its new angular speed? (The coefficient of linear expansion of brass is $\alpha$)

• A. ${\omega }_0[1+2a({\theta }_2-{\theta }_1\left)\right]$
• B. ${\omega }_0[1+a({\theta }_2-{\theta }_1\left)\right]$
• C. $\frac{{\omega }_0}{[1+2a({\theta }_2-{\theta }_1]}$
• D. None of these

Answer 1

The correct option is C $\frac{{\omega }_0}{[1+2a({\theta }_2-{\theta }_1]}$

Since no external force acts on the disc its angular momentum remain conserved:

Initial angular momentum :

$L=I\omega$

substituting the vale of I for a disk

$L_1=\left(\frac{1}{2}m{a}^2\right)\omega$ ...(1)

After temparture change sfrom ${\theta }_1$ to ${\theta }_2$

$finalradius=a(1+\alpha ({\theta }_2-{\theta }_1\left)\right)$

let new angular velocity be: $w_1$

Now angular momentum is given by :

$L_2=\frac{1}{2}m{a}^2(1+\alpha ({\theta }_2-{\theta }_1){)}^2{\omega }_1$ ..(2)

Using conservation of angular momentum $L_2=L_1$

solving $\left(1\right)=\left(2\right)$

$\omega =(1+\alpha ({\theta }_2-{\theta }_1){)}^2{\omega }_1$

using binomial approximation : $(1+x{)}^2=1+2x,whenx<<1$

${\omega }_1=\frac{\omega }{[1+2\alpha ({\theta }_2-{\theta }_1\left)\right]}$

Option $C$ is correct