of 2.95 cm is dropped on its edge onto a horizontal surface. The coin starts out with an initial angular speed of 12.1 rad/s and rolls in a striaght line without slipping.
If the rotation slows with an angular deceleration of magnitude 2.29 rad /s^2, how far does the coin roll before coming to rest (in units of m)?
2006-10-24 05:02:35 · 4 answers · asked by Mariska 2 in Science & Mathematics ➔ Physics
Given the original rotation speed and the deceleration rate, the time can be computed immediately. From this, the easiest way to figure the distance is to assume that the average speed of motion is half the original speed, and you can use the time to figure out the result quite easily.
2006-10-24 05:14:18 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Convert the deceleration to tangential deceleration = a = wr; where w = 2.29 rad/sec^2 is angular deceleration and r = d/2 = 2.95/2 cm is the radius of the dime.
Convert the initial velocity to tangential velocity = U = Wr; where W = 12.1 rad/sec is the angular velocity and r is the radius of the dime.
So we have V^2 = U^2 + 2as = 0; which says the dime will travel s = (U^2)/2a cm before coming to rest at V = 0 tangential velocity. You've already solved for U and a; so just plug in the numbers.
2006-10-24 05:35:14 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
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2016-12-28 03:44:17 · answer #3 · answered by ? 3 · 0⤊ 0⤋
radius=2.95/2=1.475cm
u=initial linear speed
=angular velocity*radius
=12.1rad/sec*1.475cm=17.545cm/sec
final velocity=0
deceleration
=angular deceleration*radius
=2.29*1.475=3.3775cm/sec^2
v^2-u^2=2*-a*s
s=u^2/2a
=17,545^2/2*3.3775
=45.57cm
=.4557m
2006-10-24 05:27:11 · answer #4 · answered by openpsychy 6 · 0⤊ 0⤋