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A uniform disk turns at 3.7 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and len...

A uniform disk turns at 3.7 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk. They then turn together around the spindle with their centers superposed. What is the angular frequency in rev/s of the combination?
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Concepts and reason

The concepts used to solve this problem are law of conservation of angular momentum and the moment of inertia.

First, determine the expressions for the moment of inertia of the disk and the system of the disk and the rod. Finally, calculate the angular frequency of the system of disk and the rod by using the law of conservation of angular momentum.

Fundamentals

The angular momentum of the disk is,

L1=I1ω1{L_1} = {I_1}{\omega _1}

Here, I1{I_1}is the moment of inertia of the disk and ω1{\omega _1}is the angular speed of the disk.

The angular momentum of the system of disk and the rod is,

L2=I2ω2{L_2} = {I_2}{\omega _2}

Here, I2{I_2}is the moment of inertia of the system of disk and rod and ω2{\omega _2}is the angular speed of the system of disk and rod.

The moment of inertia Idisk{I_{{\rm{disk}}}}of a disk of radius \ell about an axis passes through its centre is given by following expression.

Idisk=12m2{I_{{\rm{disk}}}} = \frac{1}{2}m{\ell ^2}

The moment of inertia Irod{I_{{\rm{rod }}}}of a rod of length \ell about an axis passes through its centre is given by following expression.

Irod=112m2{I_{{\rm{rod}}}} = \frac{1}{{12}}m{\ell ^2}

The moment of inertia of the disk is,

I1=12m2{I_1} = \frac{1}{2}m{\ell ^2}

Here, mmis the mass of the disk and \ell is the radius of the disk.

Given that, the length of the rod is equal to the radius ()\left( \ell \right)of the disk. So, the length of the rod is22\ell .

The moment of inertia of the system of disk and rod is,

I2=Idisk+Irod=12m2+112m(2)2=12m2+13m2\begin{array}{c}\\{I_2} = {I_{{\rm{disk}}}} + {I_{{\rm{rod}}}}\\\\ = \frac{1}{2}m{\ell ^2} + \frac{1}{{12}}m{\left( {2\ell } \right)^2}\\\\ = \frac{1}{2}m{\ell ^2} + \frac{1}{3}m{\ell ^2}\\\end{array}

From the law of conservation of angular momentum, the total initial angular momentum of the system of rod and disk is equal to the total final angular momentum of the system of rod and disk.

I1ω1=I2ω2{I_1}{\omega _1} = {I_2}{\omega _2}

Substitute12m2\frac{1}{2}m{\ell ^2}forI1{I_1}and 12m2+13m2\frac{1}{2}m{\ell ^2} + \frac{1}{3}m{\ell ^2}forI2{I_2}.

(12m2)ω1=(12m2+13m2)ω2(12)ω1=(12+13)ω2ω1=(1+23)ω2=53ω2\begin{array}{c}\\\left( {\frac{1}{2}m{\ell ^2}} \right){\omega _1} = \left( {\frac{1}{2}m{\ell ^2} + \frac{1}{3}m{\ell ^2}} \right){\omega _2}\\\\\left( {\frac{1}{2}} \right){\omega _1} = \left( {\frac{1}{2} + \frac{1}{3}} \right){\omega _2}\\\\{\omega _1} = \left( {1 + \frac{2}{3}} \right){\omega _2}\\\\ = \frac{5}{3}{\omega _2}\\\end{array}

Rearrange the above equation forω2{\omega _2}.

ω2=35ω1{\omega _2} = \frac{3}{5}{\omega _1}

Substitute 3.7rev/s3.7{\rm{ rev/s}}forω1{\omega _1}.

ω2=35(3.7rev/s)=2.22rev/s\begin{array}{c}\\{\omega _2} = \frac{3}{5}\left( {3.7{\rm{ rev/s}}} \right)\\\\ = 2.22{\rm{ rev/s}}\\\end{array}

Ans:

The angular frequency of the system of disk and the rod is2.22rev/s{\bf{2}}{\bf{.22 rev/s}}.

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