A speck of dust on a spinning dvd has a centripetal accelera - tion of 20 m/s2.

a. what is the acceleration of a different speck of dust that is twice as far from the center of the disk?

b. what would be the acceleration of the first speck of dust if the disk's angular velocity was doubled?

The centripetal acceleration of an object spinning in a circle is

a = v²/r = 20 m/s²

where

r = the distance of the object from the center of the circle

v = the tangential velocity of the object

The centripetal acceleration is also given by

a = rω² = 20 m/s²

where

ω = the angular velocity.

Part a.

If the radius is doubled while the angular velocity remains constant, then

the centripetal acceleration will double.

a = (2r) ω² = 2*20 = 40 m/s²

Answer: 40 m/s²

Part b.

If the angular velocity is doubled, then

a = r (2ω) ² = 4rω² = 4*20 = 80 m/s²

Answer: 80 m/s²

a) 40 m/s^2 b) 80 m/s^2 The formula for centripetal force is F = mv^2/r With that in mind, let's look at the questions. a. what is the acceleration of a different speck of dust that is twice as far from the center of the disk? This basically asks "What changes is r is doubled?" and looking at the formula, the effects of doubling r are two fold. First, velocity is directly related to radius, so the term v^2 is multiplied by 4 which increases the force by a factor of 4. But that's not all. The r term in the divisor is also doubled which will cause the force to be divided by a factor of 2, so the force is halved. So one half of 4 is 2. So the total force is doubled. So 20 m/s^2 * 2 = 40 m/s^2 b. what would be the acceleration of the first speck of dust if the disk's angular velocity was doubled? This is more straight forward. The v term in the formula for force is doubled. So the value of v^2 will be quadrupled. Which in turn causes the overall force to quadruple. So 20 m/s^2 * 4 = 80 m/s^2