A coin with a diameter of 2.40cm is dropped on edge on to a horizontal surface. The coin starts out with an initial angular speed of 18.0 rad/s and ro

beljuA

beljuA

Answered question

2020-10-18

A coin with a diameter of 2.40cm is dropped on edge on to a horizontal surface. The coin starts out with an initial angular speed of 18.0 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular acceleration of magnitude 1.90 rad/s, how far does the coin roll before coming to rest?

Answer & Explanation

escumantsu

escumantsu

Skilled2020-10-19Added 98 answers

Given diameter of the coin is D = 2.40cm
Radius is r=(D/2)=1.20cm
=1.2010 m2
Initial angular speed of the coin is
ωi=18.0 rads
Final angular speed is
ωf=0 rads
Angular acceleration is
α=1.90 rads2
Now we know the formula for angular displacement is
θ=ωf2ωi22α
By substituting the given values in this equation we get the value of angular displacement(θ).
After this the linear displacement is calculated by the formula
l=r(θ)

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