A circular disk of radius 2.0 m rotates, starting from rest, with a constant angular acceleration of 20.0 rad/s². What is the tangential acceleration of a point on the edge of the disk at the instant that its angular speed is 1.0 rev/s? A)40 m/s² B)79 m/s² C)110 m/s² D)120 m/s² E)zero m/s²

Question

Quizplus · Accepted Answer

We can use the following equations to solve this problem:

ω = ω0 + αt (1)
a = rα (2)

where ω is the final angular velocity, ω0 is the initial angular velocity (which is zero in this case), α is the angular acceleration, t is the time elapsed, a is the tangential acceleration, and r is the radius.

To find the time it takes for the disk to reach 1 revolution per second, we can use the formula:

ω = 2πf

where f is the frequency or number of revolutions per second.

Therefore, when ω = 2π radians/s, we have:

2π = 1 rev/s * 2π radians/rev

Solving for the time it takes:

t = ω/α = (2π/20)/2π = 0.05 s

Now, we can use equation (1) to find the final angular velocity:

ω = ω0 + αt = 0 + 20 * 0.05 = 1 rad/s

Finally, we can use equation (2) to find the tangential acceleration at the edge of the disk:

a = rα = 2.0 m * 20.0 rad/s^2 = 40 m/s^2

Therefore, the answer is B: 79 m/s^2.

A Circular Disk of Radius 2