Recent content by Vykan12

Note that V^{2}_{x} + V^{2}_{y} = V^{2}\sin^{2}(\theta) + V^{2}\cos^{2}(\theta) = V^{2} (\sin^{2}(\theta) + \cos^{2}(\theta) ) = V^{2} Using the trig identity sin^2(x) + cos^2(x) = 1

Y will be 0 since the ball is colliding with the ground.

I plotted the points on my calculator and got the linear expression a = 0.4v - 0.1333333... This seems like a strange problem to give as homework. Also, how did you come up with that expression for acceleration?

If the force applied and external forces are all constant, what gives you the impression that acceleration isn't?

I can see that's where the properties of conservative forces come into play. Interesting.

You should get F_{app} - F_{k} - mg\sin(\theta) = 0 n - mg\cos(\theta) = 0 The algebra's up to you. Edit: Yes, that's correct, assuming F is the force of gravity.

First off, you have to neglect friction since you do not know the kinetic friction coefficient of the surface. Second, the normal force doesn't cause the block to move in what you defined to be the x direction. Are you using Newton's second law for net force in the x and y directions?

Do you expect us to help you? What is giving you trouble?

Do you know how to draw a free body diagram?

My physics teacher answered this for me. The only reasons the derivations are different regarding r_i is because inertia isn't a vector whereas angular momentum is.

I've always thought that the work done by a force is the scalar product of the force vector and the potential displacement vector it would have if no other forces were acting on it. My teacher says that it is really the product of the force vector and the resultant displacement regardless of...

I think I might've found the mathematical answer to my earlier question: Going by what Ben Niehoff said, K = \frac{1}{2} Mv^{2}_{A} + \frac{1}{2} I_{A} w^{2} = \frac{1}{2} Mv^{2}_{A} + \left( \frac{1}{2} M(r_{A}w)^{2} + \frac{1}{2} I_{cm} w^{2} \right) We can use v_{cm} = r_{A}w to get...

It seems to me that you're not given enough information. If you had the initial and final velocity, you could plug into V_{f} = V_{i} + at to experimentally find the acceleration. What would be easier is to use Newton's second law (F=ma) since then you only need to know the object's mass and...

Here is how my book goes about defining the moment of inertia. Now here's how my book goes about deriving an expression for the angular momentum of a rigid body. From this the book derives that L = Iw My question is in regards to the bolded part. How come in one derivation we assume...

The velocity vector of A is going to be tangent to the circle, so its angle relative to the horizontal will be 60 degrees (you can check this with some simple geometry). We also know that the vertical compontent of this vector is 2 since it is given by object B. By pythagoras the magnitude of...