A penny is placed on the outer edge of a disk (radius = 0.151000000000000000000000 m) that rotates about an ax

A penny is placed on the outer edge of a disk (radius = 0.151000000000000000000000 m) that rotates about an axis perpendicular to the plane of the disk at its center, like a record or a CD. The period of the rotation is 1.6100000000000000 s.





(1) What, if anything, is producing the centripetal force that enables the penny to rotate along with the disk? a) Both kinetic and static friction contribute to the centripetal force. b) Only kinetic friction produces the centripetal force. c) Only static friction produces the centripetal force. d) No centripetal force is needed.





(2) The centripetal force depends on the speed at which the penny is moving. On what does this speed depend? a) Only the radius of the disk. b) Neither the period of the rotation nor the radius of the disk. c) Both the period of the rotation and the radius of the disk. d) Only the period of the rotation.





(3) What is the algebraic expression for the minimum coefficient of static friction μs necessary to allow the penny to rotate along with the disk? Express your answer in terms of the radius r of the disk, the magnitude g of the acceleration due to gravity, and the period T of the rotation.





(4) What is the minimum coefficient of static friction μs necessary to allow the penny to rotate along with the disk?

A penny is placed on the outer edge of a disk (radius = 0.151000000000000000000000 m) that rotates about an ax
I. c because the penny is fixed on the disk.


2. C again because Ac = w (squared) r


The speed of rotation is w.


3. The coefficient of friction us is Ac/g = wwr/g


where (w=2pi/T) So us= 4 Pi Pi r /(TTg)





4. This is 4 x3.14 x3.14 x 0.151 / (1.61 x 1.61 x 9.8) = 0.23