A phonograph turntable, initially rotating at 0.75 rev/s, slows down and stops in 30 s.The magnitude of its average angular acceleration for this process is: A)"1.5 rad/s²" B)"1.5$\pi$rad/s²" C)"$\pi$/40 rad/s²" D)"$\pi$/20 rad/s²" E)"0.75 rad/s²"

Question

Quizplus · Accepted Answer

We can use the formula:

Δω = αΔt

where Δω is the change in angular velocity, α is the angular acceleration, and Δt is the time interval.

At the start, the turntable is rotating at 0.75 rev/s. In 30 s, it slows down and stops, which means it goes through a change in angular velocity of:

Δω = 0 - 0.75 rev/s = -0.75 rev/s

We need to convert this to radians per second for the units to match the answer choices.

1 rev = 2π radians, so

Δω = -0.75 rev/s * 2π rad/rev = -1.5π rad/s

Substituting into the formula gives:

-1.5π rad/s = α * 30 s

Solving for α gives:

α = -1.5π rad/s / 30 s = -π/20 rad/s^2

Taking the absolute value gives:

|α| = π/20 rad/s^2 = π/20 * (1/2π) rev/s^2 = 1/40 rev/s^2

So the answer is D, 1/40 rad/s^2.

A Phonograph Turntable, Initially Rotating at 0 π\piπ Rad/s2" C) π\piπ /40 Rad/s2" D)

A Phonograph Turntable, Initially Rotating at 0 $\pi$ Rad/s²"
C) $\pi$ /40 Rad/s²"
D)