How Far Does a Rolling Coin Travel Before Stopping?

[physics1234]

A coin with a diameter of 2.20 cm is dropped on edge onto a horizontal surface. The coin starts out with an initial angular speed of 15.9 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular acceleration of magnitude 1.76 rad/s2, how far does the coin roll before coming to rest?

 

[SGT]

The formula for calculating angular velocity is similar to the one for linear velocity.
omega(t) = omega(0) + alpha*t
if you set the final angular velocity to zero, you can calculate the amount of time to stop.

 

[kyle8921]

Angular speed and velocity are two important concepts in the study of rotational motion. Angular speed refers to the rate at which an object rotates, while angular velocity takes into account both the speed and direction of rotation. In this scenario, a coin with a diameter of 2.20 cm is dropped on its edge onto a horizontal surface, with an initial angular speed of 15.9 rad/s.

As the coin rolls without slipping, its angular speed remains constant, but its angular velocity changes direction. This is because the coin is undergoing both rotational and translational motion. The angular acceleration, which is the rate of change of angular velocity, is given as 1.76 rad/s2.

To determine how far the coin rolls before coming to rest, we need to use the equations of rotational motion. The first equation relates angular acceleration to angular velocity and time:

ω = ω0 + αt

Where ω is the final angular velocity, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time taken.

In this case, we know ω0 = 15.9 rad/s, α = 1.76 rad/s2, and we want to find t when ω = 0. Solving for t, we get:

t = (0 - 15.9)/1.76 = 9.03 s

The second equation relates angular displacement to angular velocity and time:

θ = θ0 + ω0t + 1/2αt^2

Where θ is the final angular displacement, θ0 is the initial angular displacement, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time taken.

In this case, θ0 = 0, ω0 = 15.9 rad/s, α = 1.76 rad/s2, and t = 9.03 s. Solving for θ, we get:

θ = 0 + (15.9)(9.03) + 1/2(1.76)(9.03)^2 = 127.9 rad

Finally, we can convert this angular displacement to linear displacement by using the formula:

s = rθ

Where s is the linear displacement, r is the radius (in this case, half the diameter), and θ is the angular displacement.

In this case, s = (2.20/2