Circular motion question

[mikezuma3]

A 5 meter long 10Kg bar is pivoted at the center and has a 10Kg weight attached to one end of it. A man at the bottom will push the weight and release it directly below the pivot point. At what speed does it need to be going at release in order to reach the top? (g = 10 m/s2)

 

Any help will do!!

[swansont]

(Sounds like homework, so moved to HW help) What do you have so far?

[alexwang32]

What do you mean by 'reach the top'?

[Janus Sunaj]

Consider it an energy problem (not circular motion) with two situations:

(1) 10kg mass (not weight) at the bottom of its arc, with some nonzero velocity

(2) 10kg mass at the top of its arc, velocity zero

With no information about friction, assume none; therefore the sum of kinetic energy (KE) plus potential energy (PE) is constant through the arc, so KE in situation (1) must equal the increase in PE from situation (1) to situation (2).

[The young professor vik.]

Applying the theory, that the centripetal force pulling the body toward the centre equals the force due to gravity. You should use the formula; mv^2/r =mg. You can calculate v which is your velocity.

[daniton]

you can also consider it as conservation of energy problem. Just equate that the kinetic energy below is equal or greater than the change in potential energy of the body.