A compact disk player has just been turned off. A disk decelerates uniformly from an initial angular velocity of 7.90 rev/s. The disk rotates through 5.50 revolutions before coming to rest. What is the tangential acceleration of a point on the disk 5.05 cm from the axis?

To find the tangential acceleration of a point on the disk, we can use the following formula:

a = α * r

where:
a is the tangential acceleration,
α is the angular acceleration, and
r is the radius of the point from the disk's axis.

In this case, we can assume that the angular acceleration is constant since the disk decelerates uniformly. The final angular velocity is 0 rev/s, so we need to find the angular acceleration first.

Using the equation for uniformly decelerated linear motion:

vf^2 = vi^2 + 2ad

where:
vf is the final velocity (0 rev/s),
vi is the initial velocity (7.90 rev/s),
a is the acceleration, and
d is the displacement in revolutions.

Rearranging the equation, we get:

a = (vf^2 - vi^2) / (2d)

Substituting the given values, we have:

a = (0^2 - 7.90^2) / (2 * 5.50)

Now, we can calculate the angular acceleration (α).

Next, we use the formula given earlier:

a = α * r

Substituting the values, including the radius (r) being 5.05 cm:

5.05 cm = α * 5.05 cm

Simplifying the equation, we find:

α = a / r

So, the tangential acceleration of a point on the disk 5.05 cm from the axis is the same as the angular acceleration (α) we calculated earlier.