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December 18, 2014

December 18, 2014

Posted by **toshi** on Monday, November 28, 2011 at 5:10am.

- physics -
**Damon**, Monday, November 28, 2011 at 5:30amSimilar problem done for Jimmy, note different numbers for speed and radius:

geometry if a = 0 at t = 0

angle from x axis (call it a) = v t/r

Vx = -v sin a

Vy = v cos a

at start a = 0 so

V1x = 0

V1y = v

at time t

V2x = -v sin(vt/r)

V2y = v cos(vt/r)

call change d

d Vx = -v sin(vt/r)

d Vy = v[cos(vt/r)-1]

ax = -(v/t) sin(vt/r)

ay = (v/t)[cos(vt/r)-1]

so for our numbers

v/t = 5.4/.7 = 7.714

vt/r = 1.643 about 94 degrees, quadrant2

ax = -7.714 sin(1.643) = -7.694

ay= 7.714[cos(1.643)-1] = -8.270

|A| =SQRT(ax^2+ay^2) = 11.30 m/s^2

compare to v^2/r

v^2/r = 5.4^2/2.3 = 12.7 not too far off centripetal acceleration

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