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Q] A flywheel of mass 8 kg and radius 10 cm rotating with a uniform angular speed of 5 rad / sec about its axis of rotation, is subjected to an accelerating torque of 0.01 Nm for 10 seconds. Calculate the change in its angular momentum and change in its kinetic energy.​

Answer»

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  • m = 8kg
  • r = 10cm = 0.1
  • ω = 5r/s
  • τ = 0.01Nm
  • t = 10 sec

\\\large\underline{ \underline{ \sf{ \red{To\: Find:} }}} \\ \\

  • Change in ANGULAR momentum
  • Change in Kinetic energy

\\\large\underline{ \underline{ \sf{ \red{Solution:} }}} \\ \\

\boxed{\sf Torque τ = I a }

▪I = MOMENT of Inertia

I = 1/2 mr²

I = 1/2 × 8 × (0.1)²

I = 4 × 0.01

\boxed{\sf I = 0.04 kgm²}

▪ Angular acceleration, ɑ = τ/l

ɑ = 0.01 / 0.04

\boxed{ɑ = 0.25 rad/s²}

Final anglular velocity,

ω2 = ω1 + ɑt

ω2 = 5 + 0.25 × 10

ω2 = 5 + 2.5

\boxed{\sf ω2 = 7.5rad/s}

_____________________________

→ So, change in Angular momentum,

dL = L2 - L1

dL = I ( ω2 - ω1 )

dL = MR²/2 ( ω2 - ω1 )

dL = 8 × 0.1² / 2 ( 7.5 - 5 )

dL = 8 × 0.1² / 2 ( 2.5 )

dL = 8 × 0.01 /2 ( 2.5 )

dL = 0.008 /2 × 2.5

dL = 0.04 × 2.5

\boxed{\sf dL = 0.1 kgm²/s}

→ So, change in Kinetic energy,

KE = KE2 - KE1

= 1/2 Iω2² - 1/2 Iω1²

= 1/2 I (ω2² - ω1² )

= MR²/2×2 (ω2² - ω1² )

= 8 × 0.1²/4 ( 7.5² - 5² )

= 8 × 0.001 / 4 ( 56.25 - 25 )

= 0.02 × 31.25

\boxed{\sf ∆KE = 0.625J}


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