A block of wood 45 cm wide and 89 cm tall is placed on a merry-go-round which rotates at 0.88 rad/s. How far from the center may the block be placed before it will tip over? The acceleration of gravity is 980 cm/s^2.
Answer in units of cm.
This is how I tried to solve the problem:
F = mrw^2
F<= mg
mg = mrw^2
r = g /w^2
But this solution is incorrect, if you could show me step by step the right way to solve the problem, your help will be greatly appreciated!
Answer in units of cm.
This is how I tried to solve the problem:
F = mrw^2
F<= mg
mg = mrw^2
r = g /w^2
But this solution is incorrect, if you could show me step by step the right way to solve the problem, your help will be greatly appreciated!
-
r = (GX)/(W^2 Y)
X is your width
Y is your length
Frame of reference is on on the merry-go-round with the block.
you can reach this by assuming the net Torque = 0 just before the block tips, with the axis at its rear corner.
((mv^2)/r) ((1/2)Y) - ((1/2)X)(mG)=0
the (1/2)'s and m's cancle.
See you at the final on Monday Shay!
X is your width
Y is your length
Frame of reference is on on the merry-go-round with the block.
you can reach this by assuming the net Torque = 0 just before the block tips, with the axis at its rear corner.
((mv^2)/r) ((1/2)Y) - ((1/2)X)(mG)=0
the (1/2)'s and m's cancle.
See you at the final on Monday Shay!
-
nevermind the given value is omega not velocity
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