Asked by Frank
At some instant, a particle traveling in a horizontal circular path of radius 7.80 m has a total acceleration with a magnitude of 15.0 m/s2 and a constant tangential acceleration of 12.0 m/s2. Determine the speed of the particle at this instant(initial speed) and (1/8) revolution later(final speed).
Answers
Answered by
bobpursley
total acceleration^2=centacc^2 + tangentialacc^2
solve for centripetal acceleration
then v^r/r=centripacc
solve for v.
Now for later,
wf^2=wi^2+2 alpha*displacement
alpha*r=tangential acceleration, displacement=1/8 *2PI
vf^2/r^2=v^2/r^2 + 12/r * 1/8 *2PI
solve for Vf
solve for centripetal acceleration
then v^r/r=centripacc
solve for v.
Now for later,
wf^2=wi^2+2 alpha*displacement
alpha*r=tangential acceleration, displacement=1/8 *2PI
vf^2/r^2=v^2/r^2 + 12/r * 1/8 *2PI
solve for Vf
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