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## Q. 20.5

A wheel of mass M has been strengthened using a metal ring. It consists of a uniform disc of radius r and mass σ per unit area surrounded by a uniform solid ring (the rim) of radius r, negligible width, and mass 5rσ per unit length.
i) Find the mass of the two parts of the wheel in terms of M.
ii) Write down the moment of inertia of each part about the axle of the wheel.
iii) Find the kinetic energy when the wheel is rotating with an angular speed ω.
iv) Write this kinetic energy as a percentage of the kinetic energy of a hoop with the same mass and radius rotating at the same angular speed.

## Verified Solution

i)    The area of the disc is πr² , so its mass is $M_{1} = σπr²$.

The length of the ring is 2πr, so its mass is

$M_{2} = 2πr(5rσ)$

= 10 σπr² .

Hence                         M = 11 σπr²

⇒         $M_{1} = \frac{M}{11}$  and  $M_{2} = \frac{10M}{11}.$

ii) The moment of inertia of the inside disc about the axle is

$I_{2} = \frac{1}{2}M_{1}r^{2}$

$= \frac{Mr^{2}}{22}.$

The moment of inertia of the rim about the axle is

$I_{2} =M_{2}r^{2}$

$= \frac{10Mr^{2}}{11}.$

iii) The kinetic energy of the wheel is $\frac{1}{2}Iω²,$ where $I = I_{1} + I_{2}.$

$I_{1} + I_{2} = \frac{Mr^{2}}{22} + \frac{10Mr^{2}}{11}$

$= \frac{21}{22}Mr^{2}$

iv) The moment of inertia of the hoop about its axis is Mr², so its kinetic energy is $\frac{1}{2}Mr^{2}ω^{2}.$

The kinetic energy of the wheel is $\frac{21}{22} × 100%$ of the kinetic energy of the hoop, namely 95.5%. 