The angular velocity of a flywheel decreases uniformly from 1200 revol

1.	The angular velocity of a flywheel decreases uniformly from 1200 revolution per minute to 600 rpm in 5 seconds. Find the angular acceleration.
Answer» a. Initial angular velocity ω₁ = 2πn₁ Here n₁ = \(\frac {1200}{60}\) = 20 rps ω₁ = 2πn₁ = 2 × π × 20 = 40π rads^-1 Final angular velocity ω₂ = 2πn₂ Here n₂ = \(\frac {600}{60}\) = 10 rps ∴ ω₂ = 2 × π × n₂ = 2 × π × 10 = 20π rads^-1 Angular acceleration a = \(\frac {ω_2-ω_1}{t}\) = \(\frac {20π - 40π}{5}\) \(= \frac{-20π}{5}\) = -4π i.e. a = 12.57 rad s^-2. Therefore angular retardation is 12.57 rad s² .

The angular velocity of a flywheel decreases uniformly from 1200 revolution per minute to 600 rpm in 5 seconds. Find the angular acceleration.

Answer»

a. Initial angular velocity ω₁ = 2πn₁

Here n₁ = \(\frac {1200}{60}\) = 20 rps

ω₁ = 2πn₁

= 2 × π × 20 = 40π rads^-1

Final angular velocity ω₂ = 2πn₂

Here n₂ = \(\frac {600}{60}\) = 10 rps

∴ ω₂ = 2 × π × n₂

= 2 × π × 10 = 20π rads^-1

Angular acceleration

a = \(\frac {ω_2-ω_1}{t}\) = \(\frac {20π - 40π}{5}\) \(= \frac{-20π}{5}\)

= -4π

i.e. a = 12.57 rad s^-2. Therefore angular retardation is 12.57 rad s² .