Question

A coin with a diameter of 2.11 cm is dropped onto a horizontal surface. The coin starts out with an initial angular speed of 19.0 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular deceleration of 1.26 rad/s2 , how far does the coin roll before coming to rest?

Answer 1

Answer:

9.49596 m

Explanation:

$\omega_f$ = Final angular velocity = 0

$\omega_i$ = Initial angular velocity = 19 rad/s

$\alpha$ = Angular acceleration = -1.26 rad/s²

$\theta$ = Angle of rotation

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-19^2}{2\times -1.26}\\\Rightarrow \theta=143.25396\ rad$

Converting to m

$143.25396\times \pi d=143.25396\times \pi\times 0.0211=9.49596\ m$

The distance the coin rolls before it stops is 9.49596 m