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Formula:
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1)I need to find out V(\theta). But I remember that r\theta<dot>
= \omega = V(\theta)
Something seems like contradict
Where my concept wrong?
How should I...
chaoseverlasting, thanks for your reply.
I have tried to solve the problem in this way.
m_{1}v_{1}r_{1}=m_{2}v_{2}r_{2}
0.02x300x0.4 = 1.02x v_{2}x(0.4+0.5)
v_{2}=2.614m/s
angular velocity= v/r = 2.614/0.9
= 2.9 rad/s
===========================
K.E. loss = P.E. gain
1/2 mv^{2} =...
http://img215.imageshack.us/my.php?image=q1db9.jpg
I haven't any suggested answer on this problem
The most important thing is that I really don't know how to start to deal with this problem.
< I am lack of knowledge on angular momentum<<I don't know is this needed in solving this...
Thanks for your reply, I can solve the problem by this method now
1) KE = 1/2 mv_{2}
where mass is known
2) F^{c}=F^{g}
\frac{mv_{2}}{r}= mg
3)the ration between the surface of Earth and at that specific height above the surface of the earth
g is inverse proportion to r_{2}
so I can get...
1. A 100 kg satellite is placed in a circular orbit 3500 km above the surface of earth. At this elevation the acceleration of gravity is 4.09m/s_{2}. Determine the kinetic energy of the satellite , knowing that its orbitial speed is 22.9 x10_{3}km/h
The answer:30.3 GJ
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I have drawn...
You should use the formula
y=x tan theta - \stackrel{gx^2}{2u^2cos^2 theta}
THis formula is combined by
verticial component using s=ut+1/2 at^2
horizontal component v=s/t