How can I rearrange an equation for circular motion to solve for velocity?

[junkie_ball]

Hi newbie question,

I'm trying to work through an example of a equation where circular motion can be expressed . I'm having difficultly understanding where the next line in the equation comes from. I have shown them below. My goal is to express 'V' although currently i am a few steps away from this.

Line 1

mgr + 0.5m(2$\sqrt{}rg$)² = 0.5mv²

Line 2

gr + (2$\sqrt{}rg$)² / 2 = 0.5v²

I can follow the equation such that i understand both 0.5m cancels out from both sides but i get lost when i try to figure out how i lose the m from the mgr expression in line 1 from line 2?

 

[flyingsleuth]

1. original ... mgr + 0.5m(2√rg)2 = 0.5mv2

2. divide both sides by 0.5m... (mgr / 0.5m) + (0.5m(2√rg)2 / 0.5m) = v2

3. result ... (gr / 0.5) + (2√rg)2 = v2

4. rearrange ... gr + (2√rg)2 = 0.5v2

I haven't touched physics & maths for 11 years since i left university with a masters but i think that is how you get the above.

 

[junkie_ball]

1. original ... mgr + 0.5m(2√rg)2 = 0.5mv2

2. divide both sides by 0.5m... (mgr / 0.5m) + (0.5m(2√rg)2 / 0.5m) = v2

3. result ... (gr / 0.5) + (2√rg)2 = v2

4. rearrange ... gr + (2√rg)2 = 0.5v2

I haven't touched physics & maths for 11 years since i left university with a masters but i think that is how you get the above.

Thanks for the reply. It now makes sense i was forgetting about the 0.5m on the right hand side which would cancel a 0.5m on the left hand side. Thanks for the explanation. :D