Recent content by ac2707

OMG ! : you make it look so easy ! :bugeye: ok..how do you get this equation \overrightarrow {v_A } = pr\widehat i - b\omega \widehat j ? is it simply angular speed cross radius ? but why is it b\omega \widehat j ? is b the radius ?..looks funny, bcause b doesn't even touch the end of...

well..ok...did you get the answer written on the question with that equation ?..I think that's not the problem though, as I'll be using the same formula anyway thanks

I tried doing this, but I can't get the exact same answer as the one given http://img246.echo.cx/my.php?image=untitled0nj.png I use relative acceleration a_a = a_b + \dot \omega \cross r + \omega \cross (\omega \cross r) + 2 \omega \cross V_{rel} + a_{rel} and as V_{rel} = 0 \dot...

so the angular speed should be V/r ? if so, how do I espress the device acceleration ? is it a = theta(dot)^2 x r (normal axis) + theta(double dot) x r (tangential axis)?. theta as in 10 degrees.. and yes..if the accelerometer is tilted 90 degress (or vertical) it will read the acc of gravity.

well u got it right.. so the actual acceleration of the device would be V^2/r because it is on the normal axis ...but how do I determine the effect of the roll?? I think I presented the problem inaccurately at the first time..I have an image here you might want to refer to...

Ok, this is probably easy...but I'm stuck f(x,y) = y^2 + xy - x^2 +2 find the gradient of the normal to the level curve at the point (3,-2) my answer is -1/root65 , but it's supposed to be 1/8. I did it by finding Fx(x,y) = y-2x and Fy(x,y) = 2y+x then finding the absolute value of the...

I need help with this say there's a truck travellin at a constant speed along a winding flat road. the car body is connected to the axles with springs. Thus, the car body is assumed to roll at a 10degree angle from the vertical of a horizontal 'roll axis' which is fixed on the axles. An...