A body rotates along a circle of radius 0.2 m with a speed of 240 rota

1.	A body rotates along a circle of radius 0.2 m with a speed of 240 rotations per minute. Calculate angular speed linear speed If the speed of rotation changes to 330 rotations per minute in 10 seconds, calculate the angular and linear acceleration.
Answer» The angular velocity is given by ω₁ = \(\frac {2πn1}{t}\)rads^-1 Here n₁ = 240, t = 1 min = 60 s. ∴ Initial angular velocity = \(\frac{2π \times 240}{60}\)= 8π rads^-1 Initial linear velocity v₁ = rω₁; Here, r = 0.2 m ∴ v₁ = 0.20 × 8π = 5.02 ms^-1 ∴ Final angular velocity ω₂ = \(\frac{2π \times n2}{t}\) = \(\frac{2π \times330}{60}\) = 11π rads^-1 ∴ Final linear velocity v₂ = rω₂ = 0.2 × 11π = 6.91 ms^-1 From the equation of motion, we have, ω₂= ω₁ + at ∴ a = \(\frac{w_2-w_1}{t}\) Here, ω₂ = 11π, ω₁ = 8 π, ∴ α =\(\frac{w_2-w_1}{t}\) = \(\frac {11π-8π}{10}\) =\(\frac {3π}{10}\) = 0.3π rads^-2 Linear acceleration is a = rα = 0.2 × 0.3π = 0.188 ms^-2.

A body rotates along a circle of radius 0.2 m with a speed of 240 rotations per minute. Calculate angular speed linear speed If the speed of rotation changes to 330 rotations per minute in 10 seconds, calculate the angular and linear acceleration.

Answer»

The angular velocity is given by

ω₁ = \(\frac {2πn1}{t}\)rads^-1

Here n₁ = 240,

t = 1 min = 60 s.

∴ Initial angular velocity

= \(\frac{2π \times 240}{60}\)= 8π rads^-1

Initial linear velocity

v₁ = rω₁;

Here, r = 0.2 m

∴ v₁ = 0.20 × 8π

= 5.02 ms^-1

∴ Final angular velocity

ω₂ = \(\frac{2π \times n2}{t}\)

= \(\frac{2π \times330}{60}\) = 11π rads^-1

∴ Final linear velocity v₂ = rω₂

= 0.2 × 11π = 6.91 ms^-1

From the equation of motion, we have,

ω₂= ω₁ + at

∴ a = \(\frac{w_2-w_1}{t}\)

Here, ω₂ = 11π,

ω₁ = 8 π,

∴ α =\(\frac{w_2-w_1}{t}\)

= \(\frac {11π-8π}{10}\)

=\(\frac {3π}{10}\) = 0.3π rads^-2

Linear acceleration is

a = rα = 0.2 × 0.3π

= 0.188 ms^-2.