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### (GR8677 #97, #93, #75) probs from GR8677

Posted: **Thu Sep 29, 2005 2:49 am**

by **Nova**

Hello,

Classical mechanics probs from GR8677 -

#97, #93, #75

Can you people help me with these? What the heck 'synchronous' means in #75?

Thanks!

Posted: **Fri Sep 30, 2005 6:41 pm**

by **thermalEquilibrium**

75.

sync. means that the satellite moves at the same speed as the earth rotates. An example is a TV satellite, you will notice that your service dish does not move.

This problem involves two forces, centripical and gravatational, equate these and solve for v.

The period of the satellite is given by

T=2piR_sat/v=2pi(R_sat)^1.5/(GM_e)^0.5

since we are given the period at sea level, T_o=80min, we can solve for the radius of the earth, R_e,

Finally, we know the period of an sync. satellite is T=24hrs, solve for R_sat using the above eq.,

R_sat=(T/T_o)^2/3 R_e=6.8 R_e (approximate the calc in yur head)

Posted: **Fri Sep 30, 2005 6:46 pm**

by **thermalEquilibrium**

93.

find the force from the power, and then use the friction eq. F_f=muF_n

Power=VI (since its 100% efficient)

=1080J

F=Power/v=108N, this is the frictional force

the normal force is just mg

mu=F_f/F_n=VI/(mgv)=1.08

I dont have a good answer for 97, hope ive been of some help

### GR8677 #97 Solution

Posted: **Sat Oct 01, 2005 5:06 am**

by **Grant**

Thanks thermalEquilibrium. I didn't have enought time to tackle 4 problems but I couldn't say no to the one you left.

97. A conservation of angular momentum problem.

I believe somewhere in Marion & Thornton they show that the angular momentum of a rigid body is the angular momentum of the center of mass of the rigid body (relative to some point) plus the angular momentum of the rotating body about its center of mass.

Taking a positive angular momentum to be pointing out of the page you have the initial angular momentum about the point P to be:

L = -R*v0 + I*w0

(i.e. L = L_of_center_of_mass_about_P + L_about_center_of_mass)

The initial angular momentum is equal to the final angular momentum and the rest is just plugging things in.

Posted: **Sat Oct 01, 2005 7:54 pm**

by **thermalEquilibrium**

nah, no prob Grant.

Thanks alot for setting up this resource, you did a great job.

Posted: **Sun Oct 02, 2005 1:34 am**

by **Nova**

Thanks so much you guys! That was a big help. :)

I also realized #75 becomes very simple if you used Kepler's law:

a^3 = P^2, where a = radius and P = period.