A compact disk, which has a diameter of 12.0 cm, speeds up uniformly from zero to 2.50 rev/s in 2.10 s. What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is the following values?
(a) 2.00 rev/s
To find the tangential acceleration of a point on the outer rim of the disk when its angular speed is 2.00 rev/s, we need to use the formula for tangential acceleration:
a_t = r * α
Where:
a_t is the tangential acceleration
r is the radius of the disk
α is the angular acceleration
In this case, the angular acceleration can be found using the formula:
α = (ωf - ωi) / t
Where:
ωf is the final angular speed (2.50 rev/s)
ωi is the initial angular speed (0 rev/s)
t is the time taken to reach the final angular speed (2.10 s)
Substituting the values:
α = (2.50 rev/s - 0 rev/s) / 2.10 s
α = 1.19 rev/s²
Next, we can calculate the tangential acceleration using the given angular speed:
a_t = r * α
Given:
diameter of the disk = 12.0 cm
To find the radius, we divide the diameter by 2:
radius (r) = 12.0 cm / 2
radius (r) = 6.0 cm
Since the formula for tangential acceleration requires the radius to be in meters, we convert it from centimeters to meters:
radius (r) = 6.0 cm * 0.01 m/cm
radius (r) = 0.06 m
Finally, we can calculate the tangential acceleration when the angular speed is 2.00 rev/s:
a_t = r * α
a_t = 0.06 m * 1.19 rev/s²
a_t = 0.0714 m/s²
Therefore, 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 is 0.0714 m/s².