IP A 3.5-inch floppy disk in a computer rotates with a period of 2.00 \\times 10

Question

IP A 3.5-inch floppy disk in a computer rotates with a period of 2.00 imes 10^{ - 1} { m{s}}.

Part A
What is the angular speed of the disk? (Note: A 3.5-inch floppy disk is 3.5 inches in diameter.)
Express your answer using three significant figures.
omega =


m rad/s


Part B
What is the linear speed of a point on the rim of the disk?
Express your answer using two significant figures.
v =


m m/s


Part C
Does a point near the center of the disk have an angular speed that is greater than, less than, or the same as the angular speed found in part A?

greater than
less than
the same as

Explanation / Answer

d = 3.5 inch = 8.89.10-2 m

r = d/2 = 4.445.10-2 m

T = 2.00.10-1 sekon

A) = 2/T = 31.4 rad/s

B) v =.r = 1.40 m/s

C) for the point near the center of disk have same anguler speed with part A

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