Recent content by marlasca23

Thanks! My confusion was because I thought that the center of the curve was supposed to be at ground level.Thanks again :)

Sorry, I don't understand that. Could you explain it more in depth, please?

Why during a banked curve is the horizontal component of the normal force considered as the net force and the centripetal force? The horizontal component of Normal force, N·sinθ, is not even ponting towards the centre of the curvature.

The easiest way to solve this is wuth momentum(p) and impulse(J). It says the velocity is -1.47 m/s so the negative direction is away from the wall. It is important that J=Δp and p= m·v p0=0 kg·m/s J=Δp=m·Δv-0=45.1·(-1.47)=-66.297 kg·m/s J=Δp=F·Δt=-66.297 kg·m/s -66.297=F·Δt...

By the way, in case it wasn't clear, my question was already answered. How do I close the thread?

Thanks.

Yeah, I knew it was towards the center but I wanted to know the formula.

Thanks. I didn't know the position vector for uniform circular motion, but I looked it up and did the derivatives. Your advice was useful,

Recently, I was looking into centripetal acceleration and there's something I don't understand. According to my book, during uniform circular motion, the acceleration is a= v^2/r where v is the speed at which the object is moving and r is the radius of the circle. However, this formula is...