Sum

A wheel of moment of inertia 1 Kgm^{ 2 }is rotating at a speed of 40 rad/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

I = 1 kgm^{2} , c = 40 rad/s ,

t_{1} = 600s , t_{2} = 10 - 2 = 8 min = 480 s

L = ?

ω_{2} = ω_{1 }+ α t_{1}

0 = 40 + 600 α

∴ α = `(-1)/15` rad/s^{2}

ω_{3} = ω_{1 }+ α t_{2}

`= 40 - 1/15 xx 480`

∴ ω_{3 }= 8 rad/s

L = I ω_{3 }

= 1 × 8 = 8 Kgm^{2} /s

**The angular momntum of the wheel , two mintes before it comes to rest would be 8 Kgm ^{2}/s.**

Concept: Physical Significance of M.I (Moment of Inertia)

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