Question

a coin with a diameter of 2.21 cm is dropped onto a horizontal surface. the coin starts out with an initial angular speed of 17.5 rad/s and rolls in a straight line without slipping. if the rotation slows with an angular deceleration of 1.66 rad/s2 , how far does the coin roll before coming to rest? answer in units of m. answer in units of m.

Answer 1

Approximately
`1.02\; {\rm m}`.

Step-by-step explanation:

Let
`\omega_(0)` denote the initial angular velocity, and let
`\omega_(1)` denote the final angular velocity. It is given that the initial angular velocity is
`\omega_(0) = 17.5\; {\rm s^(-1)}` and
`\omega_(1) = 0\; {\rm s^(-1)}`.

Additionally, let
`a` denote the angular acceleration of this coin. It is given that
`a = (-1.66)\; {\rm s^(-2)}` (negative since angular velocity is slowing down.)

Let
`\theta` denote the angular displacement of the coin. Rearrange the rotational SUVAT equation
`({\omega_(1)}^(2) - {\omega_(0)}^(2)) = 2\, a\, \theta` to find the angular displacement
`\theta`:

`\begin{aligned}\theta &= \frac{{\omega_(1)}^(2) - {\omega_(0)}^(2)}{2\, a} \\ &= \frac{(0\; {\rm s^(-1)})- (17.5\; {\rm s^(-1)})}{2\, (1.66\; {\rm s^(-2)})} \\ &\approx 92.25 \; (\text{radians})\end{aligned}`.

It is given that the diameter of the coin is
`d = 2.21\; {\rm cm} = 0.021\; {\rm m}`.The radius of the coin would be
`r = (d/2) = ((0.021\; {\rm m}) / 2) = 0.0105\; {\rm m}`.

The question is asking for the linear displacement of this coin. Under the assumptions, the linear displacement of this coin would be equal to the product of angular displacement (in radians) and the radius of the coin:

`\begin{aligned}x &= r\, \theta \\ &= (0.0105\; {\rm m})\, (92.25) \\ &\approx 1.02\; {\rm m}\end{aligned}`.