Question

Hi! Please explain how to solve this step-by-step. I tried solving but not sure if I did it correctly. Thanks!​

Answer 1

$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: QUESTION}}}$

A flywheel used to prepare earthenware pots is set into rotation at 100rpm. It is in the form of a disc of mass 10kg and a radius 0.4m. A lump of clay (to be taken equivalent to a particle) of mass 1.6kg falls on it and adheres to it at a certain distance x from the center. Calculate x is the wheel now rotates at 80rpm

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$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: firstly \: let's \: understand \: the \:question}}}$

In the question there is given that

• A flywheel used to prepare earthenware pots is set into rotation at 100rpm.

• It is in the form of a disc of mass 10kg and a radius 0.4m.

• A lump of clay (to be taken equivalent to a particle) of mass 1.6kg falls on it and adheres to it at a certain distance x from the center.

• and we need to Calculate x is the wheel now rotates at 80rpm

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$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: Formulas \: used}}}$

$\begin{gathered}\begin{gathered}\\\;\large{\boxed{\sf{ I_1 \omega_1 \;=\; I_2 \omega_2}}}\end{gathered}\end{gathered}$

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~ Let's write what is given and what we have to find

GIVEN :-

Case 1

$\begin{gathered}\\\;\sf{\odot\;\;rotation\;=\;\bf{n_1\;=\: 100\;rpm \: = \orange{ \frac{5}{3}rps }}}\end{gathered}$

$\begin{gathered}\\\;\sf{\odot\;\;mass \: of \: disk\;=\;\bf{M \: = \orange{ 10 \: kg}}}\end{gathered}$

$\begin{gathered}\\\;\sf{\odot\;\;radius\;=\;\bf{R\: = \orange{ 0.4 \: m}}}\end{gathered}$

Case 2

$\begin{gathered}\\\;\sf{\odot\;\;rotation\;=\;\bf{n_2\;=\: 80\;rpm \: = \orange{ \frac{4}{3}rps }}}\end{gathered}$

$\begin{gathered}\\\;\sf{\odot\;\;mass \: of \: clay\;=\;\bf{m \: = \orange{ 1.6 \: kg}}}\end{gathered}$

$\begin{gathered}\\\;\sf{\odot\;\;radius\;=\;\bf{r\: = \orange{ x}}}\end{gathered}$

To FIND -

$\begin{gathered}\\\;\sf{\odot\;\;value \: of \: \bf{\green{ x}}}\end{gathered}$

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$\: \: \boxed{\boxed{\bf{\mapsto \: \: \: Solution}}}$

if something is freely falling or stuck on the rotating object law of conservation of angular momentum is applicable

$\begin{gathered}\begin{gathered}\\\;\large{\boxed{\sf{ I_1 \omega_1 \;=\; I_2 \omega_2}}}\end{gathered}\end{gathered}$

Let's find the value of $\sf I_1 ,\omega_1 ,I_2, \omega_2$

$\begin{gathered}\begin{gathered}\\\;\large{\longrightarrow{\sf{ I_1 = \frac{10 \times 0.4 \times 0.4}{2} = 0.8 \: kgm^{2} }}}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\large{\longrightarrow{\sf{ \omega_1 = 2\pi n _1= \frac{10\pi}{3} rad/s }}}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\large{\longrightarrow{\sf{ I_2 = ( 0.8 + 1.6 {x}^{2}) kgm^{2}}}}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\large{\longrightarrow{\sf{ \omega_2= 2\pi n _2= \frac{8\pi}{3} rad/s }}}\end{gathered}\end{gathered}$

Now , according to the law of conservation of angular momentum is

$\begin{gathered}\begin{gathered}\\\;\large{\boxed{\sf{ I_1 \omega_1 \;=\; I_2 \omega_2}}}\end{gathered}\end{gathered}$

~ after putting the values

$\begin{gathered}\begin{gathered}\\\;\large{\longrightarrow{\sf{ 0.8 \times \frac{10\pi}{3} = (0.8 + 1.6 {x}^{2} ) \frac{8\pi}{3} }}}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\large{\longrightarrow{\sf{ 1 = 0.8 + 1.6 {x}^{2} }}}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\large{\longrightarrow{\sf{ 1 .6 {x}^{2} =( 1 - 0.8 = 0.2) }}}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\large{\longrightarrow{\sf{ \therefore{x}^{2} = \frac{0.2}{1.6} = \frac{1}{8} }}}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\large{\longrightarrow{\sf{ {x}= \frac{1}{ \sqrt{8} } = \frac{1}{ \sqrt[2]{2} } }}}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\large{\longrightarrow{\sf{ {x}= \frac{0.707}{{2} } }}}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\large{\Longrightarrow{\sf{ {x}= {0.353 \: m} }}}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\\\;\underline{\boxed{\tt{\odot\;\;Hence,\;\;value \: of \;x \;=\;\bf{\blue{0.353 \: m}}}}}\end{gathered} \end{gathered}\end{gathered}\end{gathered}$

Answer 2

Question :

A flywheel used to prepare earthenware pots is set into rotation at 100 rpm. It is in the form of a disc of mass 10kg and radius 0.4 m. A lump of clay (to be taken equivalent to a particle) of mass 1.6 kg falls on it and adheres to it at a certain distance x from the centre. Calculate x if the wheel not rotates at 80 rpm.

Given :

Mass of flywheel = 10 kg

Radius of flywheel = 0.4 m

Mass of particle = 1.6 kg

Initial frequently = 100 rpm

Final frequency = 80 rpm

Particle is revolving in the circle of radius x m.

To Find :

The value of x.

Solution :

❖ Since no external torque acts on the whole system, angular momentum is conserved.

• We know that angular momentum is measured as the product of moment of inertia and angular velocity.

Mathematically, L = I ω

❖ Moment of inertia of a disc about an axis passing through centre and perpendicular to its plane is given by,

➙ I₁ = MR²/2

➙ I₁ = 10 × (0.4)² / 2

➙ I₁ = 5 × 0.16

➙ I₁ = 0.8 kg m²

$\sf:\implies\:L_{Initial}=L_{Final}$

$\sf:\implies\:I_1\omega_1+I_2\omega_2=(I_1+I_2)\omega$

; where ω denotes angular velocity of the combined mass

• Initial angular momentum of particle is zero.

$\sf:\implies\:(0.8)(2\pi \nu_1)+0=(0.8+I_2)(2\pi \nu_2)$

; where $\nu$ denotes frequency of rotation

$\sf:\implies\:0.8\times 100=(0.8+I_2)\times 80$

$\sf:\implies\:0.8+I_2=\dfrac{80}{80}$

$\sf:\implies\:I_2=1-0.8$

$\bf:\implies\:I_2=0.2\:kg\cdot m^2$

♦ Moment of inertia of particle :

$\sf:\implies\:I_2=mx^2$

$\sf:\implies\:0.2=1.6\times x^2$

$\sf:\implies\:x^2=0.125$

$\sf:\implies\:x=\sqrt{0.125}$

$:\implies\:\underline{\boxed{\bf{\orange{x=0.353\:m=35.3\:cm}}}}$