Recent content by pff

I understand now that i would be ignoring the direction of the acceleration by subtracting after taking the sum. If i could subtract the 9.81 from the z, could i then go ahead with summing the accelerations and then integrating? Is this be the same as integrating separately?

Thanks for the reply, but the idea of the centrefuge has confused me even more. Wouldn't we get the same result by summing before integrating rather than after? (comparing the two methods i outlined in the first post) Treating x y and z independantly gives me a major headache in that I...

I'm trying to integrate acceleration from an acceleraometer to find a distance travelled. I have heard all the stories about this not being accurate but i didn't come up with the method I'm just trying to implement an algorithm to do it. I'ts justan estimate for wear rates, not positioning...

ok, thanks guys now i get what torque I am supposed to put in the equation, but i don't know how i would work that out without doing it backwards using the energy given to the ball and then finding the torque. I suppose you need to know the the energy transferred to the ball to know the torque...

I thought torque was supposed to be a force, if its rate of change of angular speed then its an angular acceleration? so i can't just say the torque on the ball is the same as the stall torque of the drive motor? is the original equation right then, and i just need a different value for...

hi, i was given \frac{1}{2}mv^{2} = 2\tauΔ\theta for working out kinetic energy transferred to a ball in a tennis ball launcher type scenario, where two wheels are counterrotating and the ball goes between them. Is this right? is the speed of the wheels not taken into account? If the...

woohoo! Thanks for your time and patience, its very much appreciated.

Ff = (2/5)*m*a ma = m*g*sin(t) - Ff ma = m*g*sin(t) - (2/5)*m*a a = g*sin(t) - (2/5)*a a + (2/5)*a = g*sin(t) (7/5)*a = g*sin(t) a = g*sin(t) / (7/5) ..? please?

t = I*a/r = (2/5)*m*r^2*a / r t = Ff*r Ff*r = (2/5)*m*a*r F = (2/5)*m*a the force? that's what we want first right?

ok, but if i have t = I\alpha so I = (2/5)*m*r^2 but angular acceleration is due to the angular component of gravity, although the friction causes the rolling, the amount is due to g isn't it? ... unless i use a = \alphar \tau = I*a/r is that better?

sorry, so i have F = m*g*sin(angle) - F(friction) and t = rF = r*m*g*sin(angle) ...?

Sorry, a = g*sin(angle) - F / m I thought torque from friction was equivalent to the force pulling the ball down the slope in order to make it rotate torque = m*g*sin()*r alpha = a/r, but i don't understand where that gets me, back to a = g*sin(angle) / (2/5) ?

I guess i have to account for friction then? a = m*g*sin(angle) - F / m Your saying apply Newtons laws twice? i suppose this is to get a net force? F(tran) = m*g*sin(angle) - F(friction) F(rot) = ma = m*g*sin(angle)/(2/5) F = m*g*sin(angle) - F(friction) + m*g*sin(angle)/(2/5)...

well i already have translational acceleration a = g*sin(angle) ...can i just add them together? a = a(rot) + a(tran) = g*sin(angle) / (2/5) + g*sin(angle)

Translational motion due to the ball slipping? Now my head is starting to hurt. I would need to have a value for friction, then split the force between rotational acceleration and acceleration down the plane? I don't have any specific value for friction coefficient. If there is sufficient...