A phonograph record accelerates from rest to 41.0 rpm in 4.73 s. (a) What is its angular acceleration in rad/s2? rad/s2 (b) How many revolutions does it go through in the process? rev

(a) α = 0.908 rad/s²

(b) Ф = 1.62 rev.

Explanation:

(a)

Using,

α = (ω-ω₀)/t Equation 1

Where α = angular acceleration, ω = final angular velocity, ω₀ = Initial angular velocity, t = time.

Given: ω₀ = 0 rad/s (from rest), ω = 41 rpm = 41(0.10472) = 4.29352 rad/s, t = 4.73 s

Substitute into equation 1

α = (4.29352-0)/4.73

α = 0.908 rad/s²

(b)

using,

Ф= (ω+ω₀)t/2Equation 2

Where Ф = revolution in rad.

Given: ω₀ = 0 rad/s. ω = 4.29352 rad/s, t = 4.73 s

Substitute into equation 2

Ф = (4.29352+0)4.73/2

Ф = 10.154 rad

Ф = 10.154/2π = 10.154/(2×3.14)

Ф = 10.154/6.28

Ф = 1.62 rev.

the correct choice is a .

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