Recent content by wolf party

right i see, wikid - for |F(cor)| i get 2*75kg*(0.52rad/s *1m/s) = 78N thanks for your help

but yeah youve made sense of the first part and yeah i was getting mixed up with tangential speed,thanks for that

how can w be out of the page when it is rotating anti-clockwise? w is angular velocity?

my calculation shows that the astronauts weight is larger when he is running with the angular rotation of the cylinder, and lighter when he is running against it. is this correct? because it doesn't make logical sense to me. the next part deals with coriolis force. the astronaut climbs a...

thanks for that, think I've got it nailed

to be fair I am not sure what i mean by it also! so for an astronaut standing in this wheel his apparent weight is just mw2r - that makes enough sense - but i i really don't know how this changes when he is running along the wheel - all i have been told is that it is easier to run one way...

Homework Statement we have a spaceship with a spinning wheel artificial gravity thingy on-board with radius 25m, period ~ 12s, a=6.87 it is spinning anticlockwise if we take a perspective looking down on it (in the bottomleft is point A,bottom right point B,and at the top point C) an...

anyone?

Homework Statement a light source a(x) is defined by a(x) = Acos(pi*x/a)[theta(x+(a/2)) -theta(x-(a/2))] calculate the diffraction pattern I(X) Homework Equations I(X)=2pi|a~((2pi/(LAMBDA*d))*X)|2 this is the equation for a diffraction pattern on a screen at distance d from...

FNet = N - mg = 3/4mg therfore N = 1/4mg ?

Fnet = mg + mgl/2 ?

the force at the CM=3/4*mg Net force at pivot = force at CM ?

acceleration at CM = 3/4*g

is the acceleration ofthe CM just 3TAU/2mL again the reacton force at the pivot and mg at the CM are the forces on the wood R = 0 because mgl/2 (where l = 0) = 0 therfore ma=mg ?

so the acceleration due to gravity is a = 3*TAU/2*ML at the CM and the torque about the pivot is mgl/2 so m*a = TAU*(1/2*l) ?