A coin with a diameter of 2.10 cm is dropped on edge onto a horizontal surface. The coin starts out with an initial angular speed of 13.3 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular acceleration of magnitude 2.21 rad/s2, how far does the coin roll before coming to rest?
To find the distance the coin rolls before coming to rest, we need to find the time it takes for the coin to stop rotating.
We can use the rotational motion equation: ωf = ωi + αt, where
ωf is the final angular velocity (which is 0 since the coin comes to rest),
ωi is the initial angular velocity (13.3 rad/s),
α is the angular acceleration (-2.21 rad/s^2),
and t is the time.
Setting ωf to 0 and solving for t, we get:
0 = 13.3 rad/s + (-2.21 rad/s^2) * t
Rearranging the equation, we have:
2.21t = 13.3
Dividing both sides by 2.21, we find:
t = 13.3 / 2.21
t ≈ 6.01 seconds
Now that we know the time it takes for the coin to stop rotating, we can find the distance the coin rolls using the equation: d = v_avg * t,
where v_avg is the average linear velocity of the coin.
The average linear velocity of the coin is the product of the radius of the coin and the angular velocity:
v_avg = r * ω_avg
The radius of the coin is half of its diameter, so:
r = 2.10 cm / 2 = 1.05 cm = 0.0105 m
The average angular velocity can be found by taking the average of the initial and final angular velocities, which is 0 rad/s in this case since the coin comes to rest.
ω_avg = (ωi + ωf) / 2 = (13.3 rad/s + 0 rad/s) / 2 = 6.65 rad/s
Substituting the values into the equation, we have:
d = (0.0105 m) * (6.65 rad/s) * (6.01 s)
Calculating the value, we find:
d ≈ 0.3969 meters
Therefore, the coin rolls approximately 0.3969 meters before coming to rest.