Asked by kwaku
C) A steel ball bearing the radius 1 cm is rolling along a table with speed 20 cm/sec when it starts to roll up an incline. How high above the table level will it rise before stopping? Ignore friction losses
Answers
Answered by
drwls
The kinetic energy of a rolling solid sphere is
(1/2)MV^2 + (1/2)I w^2
= (1/2)M*V^2 + (1/2)(2/5)M(R*w)^2
= (7/10)M*V^2
Set that equal to M*g*H and solve for the height that it rises, H.
H = (0.7) V^2/g
The radius doesn't matter. Neither does the material.
It will only rise about 1.1 cm
20 cm/s is quite slow
(1/2)MV^2 + (1/2)I w^2
= (1/2)M*V^2 + (1/2)(2/5)M(R*w)^2
= (7/10)M*V^2
Set that equal to M*g*H and solve for the height that it rises, H.
H = (0.7) V^2/g
The radius doesn't matter. Neither does the material.
It will only rise about 1.1 cm
20 cm/s is quite slow
Answered by
Jd
it actually equals 0.28 cm I have no clue how you got 1.1 cm
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