Question

Angular speed of a particle increases from 2rads to 4 rads across any two diametrically opposite positions. Its angular acceleration will be?
A) 6 rad s?
6
-rad s-2
rad 3
rad s-2
6
Di rad s-2​

Answer 1

Answer :-

• Angular displacement of the particle after 4 s, θₜ = 32 rad

Explanation :-

Given :-

• Initial angular speed of the particle, ω₀ = 2 rad/s
• Angular acceleration of the particle, α = 3 rad/s²
• Time, t = 4 s

To find :-

• Angular displacement of the particle after 4 s, θₜ = ?

Knowledge required :-

Equation for angular displacement :

⠀⠀⠀⠀⠀⠀⠀⠀θₜ = ω₀t + ½αt²⠀

[Where : θₜ = Angular displacement of the particle, ω₀ = Initial angular speed of the particle, α = Angular acceleration of the particle and t = Time taken]

Solution :-

• To find the angular displacement of the particle after 4 s :

• By using the formula for angular displacement of a particle and substituting the values in it, we get :

• ⠀⠀=> θₜ = ω₀t + ½αt²
• ⠀⠀=> θ₄ = 2 × 4 + ½ × 3 × 4²
• ⠀⠀=> θ₄ = 8 + ½ × 3 × 16
• ⠀⠀=> θ₄ = 8 + 24
• ⠀⠀=> θ₄ = 32

⠀⠀⠀⠀⠀∴ θ₄ = 32 rad

Hence, the angular displacement of the particle after 4 s is 32 rad.

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