Asked by ruby


A coin with a diameter of 3.30 cm is dropped on edge onto a horizontal surface. The coin starts out with an initial angular speed of 16.1 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular acceleration of magnitude 1.93 rad/s2, how far in meters does the coin roll before coming to rest?

that was the question and i didn't understand it so you told me to show you my work. here it is...


wi = 16.1 rad/s = 2.5624 rev/s
a = -1.93 rad/s^s = -.3072 rev/s^2


0=2.6^2+2(.3072)theta
theta=11.0026 rev
2pi*r*theta

and i get the answer to be 1.14m but that's not right.
Answered by bobpursley
Hmmmm, It appears to me you did too much rounding. I do not understand all the conversions .


wf^2=Wi^2+2ad

0=16.1^2+2(-1.93)d

d= 16.1^1/3.86 radians

distance= angdisplacement*r
= 16.1^2/3.86 * .0330/2=1.11 m

check that.
Answered by ruby
yes, Thanks! but... i don't understand how you got the .0330 and why you divide the whole thing by 2.
Answered by ruby
wait. nevermind about the .0330. but why divide by 2?
Answered by Damon
diameter of 3.30 cm /100 = .033 meters

but we use w r, not w D
so divide diameter by two to get radius
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