Help! Just can't get this one.?

Question

A coin that has a diameter of 2.95is dropped on edge onto a horizontal surface.?
The coin starts out with an initial angular speed of 12.1 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular deceleration of magnitude 2.29 rad/s^2, how far does the coin roll before coming to rest (answer in m)?

Answer 1

Angular kinematics is very similar to straight-line-movement kinematics, i.e.
Straight: F=M*A – 2nd Newton
Angular: T=J*w’ – 2nd Newton as well for Torque T & angular acceleration w’ & so on!
Now your problem: if coin were sliding on its side with start velocity V=12.19m/s and its deceleration were a=2.29m/s^2, then you would write the following equations:
The path the coin slid before coming to a rest S=a*t^2/2 & V=a*t, isn’t it?
t=V/a hence S=a*(V/a)^2/2, so S=V^2/(2*a)
Now to angular terms! U=w’*t^2/2 & W=w’*t, where U is angular path, w’ is angular deceleration & W is angular velocity of the coin. Thus U=W^2/(2*w’). If the coin did 1 rotation & stopped, then its path would be L=pi*D. The path of our rolling coin L=U*D
Thus L=D*W^2/(2*w’)

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Answer 2

The dia of the coin 2.95 m (unit not provided so assumed)
Initial Linear velocity u=radius of coin x angular velodity (rad/s)
u=2.95x12.1
u=35.7 m/s

linear deceleration=radius of coin x radial deceleration(rad/s^2)
a=2.95 x 2.29
a=6.7555 m/s^2

Now we have the equation V^2=U^2+2as
v- final velocity m/s
u-initial velocity m/s
a-acceleration m/s^2
s-distance m

let us say that the final velocity becomes zero when the coin stops rolling.
0^2=35.7^2+2(-6.7555).s

solve for s
s=94.32 m

so the coin rolls up to 94.32 meters