A wheel of moment of inertia $1 {k g m}^{2}$ is rotating at a speed of $40 r a d / s$ . Due to friction on the axis, the wheel comes to rest in $10$ minutes. Calculate the angular momentum of the wheel, two minutes before it comes to rest.

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Solution

Given : $I = 1 k g m^{2}$
$ω_{1} = 40 r a d / s$ , $ω_{2} = 0$ at
$t = 10$ minutes $= 60 \times 10 s = 600 s$
$t^{'} = 8$ minutes $= 60 \times 8 s = 480 s$
$ω_{2} = ω_{1} + α t$
$α = \frac{ω_{2} - ω_{1}}{t}$
$= \frac{0 - 40}{600} = - \frac{1}{15} r a d / s^{2}$
At time $t^{'}$ $ω_{3} = ω_{1} + α t^{'}$
$= 40 - \frac{1}{15} \times 480$
$= 40 - 32 = 8 r a d / s$
$∴$ $L = I ω_{3}$
$= I \times 8 = 8 k g m^{2} / s$
This is the required angular momentum of the wheel.

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A wheel of moment of inertia 1kgm2 is rotating at a speed of 40rad/s. Due to friction on the axis, the wheel comes to rest in 10 minutes. Calculate the angular momentum of the wheel, two minutes before it comes to rest.

Given : I=1kgm2 ω1=40rad/s, ω2=0 at t=10 minutes =60×10s=600s t′=8 minutes =60×8s=480s ω2=ω1+αt α=ω2−ω1t =0−40600=−115rad/s2At time t′ ω3=ω1+αt′ =40−115×480 =40−32=8rad/s∴ L=Iω3 =I×8=8kgm2/sThis is the required angular momentum of the wheel.

A wheel of moment of inertia $1 {k g m}^{2}$ is rotating at a speed of $40 r a d / s$ . Due to friction on the axis, the wheel comes to rest in $10$ minutes. Calculate the angular momentum of the wheel, two minutes before it comes to rest.