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?
Adding onto bobpursley (and I know this was 11 years ago):
Wf^2=Wi^2+2ad
0=16.1^2-2*1.93d
-259.21 = -3.86d
67.1528 radians = displacement
distance= displacement*radius
So distance = 67.1528 * (3.3/2)
Distance = 40.6987 cm
To convert to m divide by 100 so
40.6987/100 = .406987 m
The coin rolls about .406987 m before coming to rest.
Wf^2=Wi^2+2a*d where d is what you are looking for.
0=16.1^2-2*1.93 d and displacement is in radians.
Then distance= displacementinradians*diameter.
To bobpursley: Your method in doing the question is correct if instead of multiplying with the diameter, you use the radius instead.
w = Wi - 1.93 t
w = 0 at stop
1.93 t = 16.1
t = 16.1/1.93
then
theta = Wi t + (1/2) a t^2
theta = 16.1 t - (1/2)(1.93) ( t )^2
then
distance = R theta
= (1.65/100) theta
i keep getting the wrong answer... still.
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.