Question

A wheel of diameter 1.0 m is rotating about a fixed axis with an initial angular
speed of 2rev s–1. The angular acceleration is 3 rev s–2.
(a) Compute the angular velocity after 2 seconds.
(b) Through what angle would the wheel turned during this time?
(c) What is the tangential velocity of a point on the rim of the wheel at t =
2 s?
(d) What is the centripetal acceleration of a point on the rim of the wheel at
t = 2 s?

Answer 1

A wheel of diameter 1.0 m is rotating about a fixed axis with an initial angular

speed of 2rev s–1. The angular acceleration is 3 rev s–2.

(a) using formula,

ω = ω0 + αt

here, ω0 = 2rev/s , α = 3rev/s² and t = 2s

so, ω = 2 + 3 × 2 = 8 rev/s

(b) using formula,

θ = ω0t + 1/2 αt²

= 2 × 2 + 1/2 × 3 × (2)²

= 4 + 6

= 10rev

we know, 1 revolution = 2π rad

so, θ = 10 × 2π = 20π rad

(c) tangential velocity at( t = 2s), v = r × ω

= 0.5 m × 8 rev/s

= 0.5 m × 8 × 2π rad/s

= 8π rad/s

(d) centripetal acceleration at t = 2s, a = v²/r

= (8π)²/(0.5)

= 128π² rad/s²

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Answer 2

$\huge\bold\purple{Answer:-}$

(a) using formula,

ω = ω0 + αt

here, ω0 = 2rev/s , α = 3rev/s² and t = 2s

so, ω = 2 + 3 × 2 = 8 rev/s

(b) using formula,

θ = ω0t + 1/2 αt²

= 2 × 2 + 1/2 × 3 × (2)²

= 4 + 6

= 10rev

we know, 1 revolution = 2π rad

so, θ = 10 × 2π = 20π rad

(c) tangential velocity at( t = 2s), v = r × ω

= 0.5 m × 8 rev/s

= 0.5 m × 8 × 2π rad/s

= 8π rad/s

(d) centripetal acceleration at t = 2s, a = v²/r

= (8π)²/(0.5)

= 128π² rad/s²