Question

A uniform circular disc with its plane horizontal is rotating about a vertical axis passing through its centre at a speed of 180 r.p.m. A small piece of wax of mass. 1.9 g falls vertically on the disc and sticks to it at a distance of 25 cm from the axis. If the speed of rotation is now reduced by 60 r.p.m., calculate moment of inertia of the disc. (Ans: 2.374 x 10⁻⁴ kgm²)

Answer 1

given, $n_1=180rpm$
so, $n_1=180/60 r.p.s=3r.p.s$

$n_2=(180-60)rpm=120rpm$
so, $n_2=120/60r.p.s=2r.p.s$

mass of wax, m = 1.9 × 10^-3 kg

now from conservation of angular momentum, because there is no external torque acts on system, angular momentum is conserved.

e.g., initial angular momentum = final angular momentum
$I_1\omega_1=I_2\omega_2$
Here, $I_2=I_1+I_{wax}$[ from parallel axis theorem ]
so, $I_2=I_1+mr^2$

now, $I_1\omega_1=I_2\omega_2$
$I_1(2\pi n_1)=(I_1+mr^2)(2\pi n_2)$

$I_1=\frac{mr^2n_2}{n_1-n_2}$

= {1.9 × 10^-3 × (0.25)² × 2}/(3 - 1)

= 1.9 × 10^-3 × 0.0625 × 2

= 3.8 × 10^-3 × 0.0625 kgm²

= 2.374 × 10^-4 kgm²

Answer 2

Answer:

The moment of inertia of the disc is $I_{1}=2.374\times10^{-4}\ Kgm^2$.

Explanation:

Given that,

Angular speed = 180 r.p.m

Mass of wax = 1.9 g

Distance d = 25 cm

After reduce,

Angular speed = 60 r.p.m

Number of revolution

$n_{1} = \dfrac{180}{60}=3\ r.p.s$

After reduce,

Number of revolution

$n_{2}=\dfrac{180-60}{60}=2\ r.p.s$

The moment of inertia is the product of mass and square of radius of the disc.

Using conservation of angular momentum

Initial angular momentum = Final angular momentum

$I_{1}\omega_{1}=I_{2}\omega_{2}$.....(I)

According to parallel axis theorem

$I_{2} =I_{1}+I_{wax}$

So, $I_{2}=I+mr^2$

Put the value of I₂ in equation (I)

$I_{1}\omega_{1}=I_{2}\omega_{2}$

$I_{1}\omega_{1}=(I_{1}+mr^2)\omega_{2}$

$I_{1}2\pi n_{1}=(I_{1}+mr^2)2\pi n_{2}$

$I_{1}=\dfrac{mr^2n_{2}}{n_{1}-n_{2}}$

$I_{1}=\dfrac{1.9\times10^{-3}\times(0.25)^2\times2}{3-2}$

$I_{1}=0.0002374\ Kgm^2$

$I_{1}=2.374\times10^{-4}\ Kgm^2$

Hence, The moment of inertia of the disc is $I_{1}=2.374\times10^{-4}\ Kgm^2$.