A compact disk, which has a diameter of 12.0 cm, speeds up uniformly from zero to 4.40 rev/s in 3.20 s.
What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00(rev/s)?
To find the tangential acceleration of a point on the outer rim of the disk, we can use the formula:
Tangential acceleration = Radius × Angular acceleration
First, let's find the radius of the disk. The diameter of the disk is given as 12.0 cm, so the radius (r) can be calculated by dividing the diameter by 2:
r = diameter/2 = 12.0 cm/2 = 6.0 cm = 0.06 m
Next, we need to find the angular acceleration of the disk.
We are given that the disk speeds up uniformly from zero to 4.40 rev/s in 3.20 s. The angular velocity (ω) is given by:
ω = Δθ/Δt
where Δθ is the change in angular displacement and Δt is the change in time. In this case, the initial angular velocity is zero, and the final angular velocity is 4.40 rev/s. Thus, the change in angular velocity (Δω) is:
Δω = final angular velocity - initial angular velocity
= 4.40 rev/s - 0 rev/s
= 4.40 rev/s
The change in time (Δt) is given as 3.20 s.
Thus, the angular acceleration (α) can be calculated using the formula:
α = Δω/Δt
= 4.40 rev/s / 3.20 s
≈ 1.38 rev/s^2
To convert the angular acceleration to radians per second squared, we know that 1 revolution (rev) is equal to 2π radians. Therefore, the angular acceleration (α) in radians per second squared can be calculated as:
α = 1.38 rev/s^2 × 2π rad/rev
≈ 8.68 rad/s^2
Finally, using the formula for tangential acceleration, we can calculate:
Tangential acceleration = Radius × Angular acceleration
= 0.06 m × 8.68 rad/s^2
≈ 0.52 m/s^2
Therefore, the tangential acceleration of a point on the outer rim of the disk, when its angular speed is 2.00 rev/s, is approximately 0.52 m/s^2.