A coin with a diameter of 1.37 cm is dropped
on its edge onto a horizontal surface. It
starts rolling with an initial angular speed
of 10.5 rad/s and rolls in a straight line without
slipping.
If the rotation slows with an angular deceleration
of 2.34 rad/s
2
, how far does the coin
roll before coming to rest?
Answer in units of m.
V^2 = Vo^2 + 2ax
0 = ((10.5rad/s)(0.00685m))^2 + (2)(2.34rad/s^2)(0.00685m)(x)
x = 0.16137 meters
To find the distance the coin rolls before coming to rest, we can use the concept of rotational motion. The key information required for this calculation is the initial angular speed, angular deceleration, and the diameter of the coin.
Given:
- Diameter of the coin = 1.37 cm = 0.0137 m
- Initial angular speed (ωi) = 10.5 rad/s
- Angular deceleration (α) = -2.34 rad/s^2 (negative sign indicates deceleration)
Using the equation for angular motion:
ωf^2 = ωi^2 + 2αθ
where:
- ωf is the final angular speed (0 rad/s, since the coin comes to rest)
- ωi is the initial angular speed (10.5 rad/s)
- α is the angular deceleration (-2.34 rad/s^2)
- θ is the angular displacement (unknown in this case)
We can rearrange the equation to solve for θ:
θ = (ωf^2 - ωi^2) / (2α)
Substituting the known values:
θ = (0^2 - (10.5 rad/s)^2) / (2 * -2.34 rad/s^2)
Calculating the value of θ:
θ = (0 - 110.25) / -4.68
θ = 23.55 rad
Now, we can find the distance rolled (d) using the formula:
d = θ * r
where:
- d is the distance rolled
- θ is the angular displacement
- r is the radius of the coin (which is half of the diameter)
Substituting the known values:
d = 23.55 rad * 0.0137 m / 2
Calculating the value of d:
d ≈ 0.0809 m
Therefore, the coin rolls approximately 0.0809 meters before coming to rest.