I think this now is the correct answer.
Final moment of inertial (if)= 250kgm^2+(35kg)(2m)^2=390kgm^2
IiWi=IfWf
(250)(15)=390wf
3750=390wf
= 9.62 rev/min
If it was a snake it would have bitten me.
So G is 6.674 x 10^-11N (m/kg)^2 correct?
That would mean my equation is:
F= 6.674 x 10^-11(50.0kg)(500kg-200kg)/0.25m^2 =1.60176 x 10^-5N toward 500kg?
How does this look?
Ok so moment of intertia =35kg(2^2)=140
15rev/m*pi/30=1.57 rad/s
Linitial=Lfinal
Linitial=I*angular speed
250(1.57)=263.12w
392.5=263.12w
1.49 rad/s=w
Converting back to rev/min:
(1.49*60)/2pi=14.23 rev/min
I have:
F= Gm1m2/r^2
F= G(50.0kg)(500kg-200kg)/0.25m^2 =240000N toward 500kg
I feel that I have it setup right but for some reason when I calculate I get 2400000 which seems wrong to me.
Confused :(
Homework Statement
A 200 kg object and a 500 kg object are separated by 0.500 m. Find the net gravitational force exerted by these objects on a 50.0 kg object placed midway between them.
What do I need to do with this problem?
Homework Statement
A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kgm2 and is rotating at 15.0 rev/min about a frictionless vertical axle. Facing the axle, a 35.0 kg child hopes onto the merry-go-round. What is...