Question

A coin with a diameter of 2.1 cm is dropped on edge onto a horizontal surface. The coin starts out with an initial angular speed of 15 rad/s and rolls in a straight line without slipping. The rotation slows with an angular acceleration of magnitude 1.7 rad/s$^2.
a) What is the initial linear velocity of the coin
b)What is the magnitude of the linear acceleration?
c)How far does the coin roll before it stops?

Answer 1

solution

given data-

initial angular velocity($\omega_0$ )=15 rad/s

radius of coin(R)=2.1/2 =1.05 cm

angular acceleration($\alpha$ )=1.7 rad/s^2

(a)initial linear velocity can be given as for pure rotation

$v=\omega _0R$

then

$v=15\times1.05\times10^{-2}=$ $0.16 m/s$

(b) linear acceleration can be written as

$a=\alpha R$

then

$a=1.7\times1.05\times10^{-2}=0.02m/s^2$

(c)

by applying equation of motion for rotation

$\omega^2=\omega_0^2-2\alpha \theta$

when coin stop then $\omega=0$

then

$\theta=\frac{\omega_0^2}{2\alpha}=\frac{15^2}{2\times 1.7}=66.18rad$

distance traveled by coin

$s=\theta R \\s=66.18\times 1.05\times 10^{-2}\\s=0.6949m\\ and \\s=0.7m$

so distance rolled by coin before stop is 0.7 m.