Asked by Anon
A bowling ball is rolled up an incline. If the ball has a speed of 12 m/s at the base of the incline, to what height h on the incline will the center mass travel?
Answers
Answered by
bobpursley
mgh=1/2 mv^2
hg=1/2 v^2
h= v^2/2g
hg=1/2 v^2
h= v^2/2g
Answered by
Anon
I tried that and did not get the right answer.
I think I need to use KE_i + PE_i = KE_f + PE_f ... which would be:
.5mv_i^2 + 1/2Iw_i^2 + mgh_i = 1/2mv_f^2 + .5Iw_f^2 +mgh_f
--> I= 2/5mr^2 for a solid sphere
I factored out the mass on both sides but my problem is that I need to convert it to an angular speed using w=v/r ... BUT I was not given a radius
I also don't have a mass for when I plug I into the equation above
I think I need to use KE_i + PE_i = KE_f + PE_f ... which would be:
.5mv_i^2 + 1/2Iw_i^2 + mgh_i = 1/2mv_f^2 + .5Iw_f^2 +mgh_f
--> I= 2/5mr^2 for a solid sphere
I factored out the mass on both sides but my problem is that I need to convert it to an angular speed using w=v/r ... BUT I was not given a radius
I also don't have a mass for when I plug I into the equation above
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