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?

Question

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.

bobpursley · Answer

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.

ruby · Answer

yes, Thanks! but... i don't understand how you got the .0330 and why you divide the whole thing by 2.

ruby · Answer

wait. nevermind about the .0330. but why divide by 2?

Damon · Answer

diameter of 3.30 cm /100 = .033 meters but we use w r, not w D so divide diameter by two to get radius