Tag Info

New answers tagged


This solution uses multiple integrals but you don't need to compute them. The final computation is a single integral. The ball has spherical symmetry so its momentum of inertia is the same with respect to the $x$-axis, the $y$-axis and the $z$-axis $I=I_x=I_y=I_z$ with $$I_z=\iiint \rho(r)\left(x^2+y^2\right)\mathrm dx\mathrm dy\mathrm dz$$ now sum up the ...


As you can see, $\rho$ only depends on a single variable: $r$. Thus, it should be intuitive that one can do this problem by integrating only over the variable $r$. To see what you are supposed to do, consider what happens if you fix $r$: You obtain a spherical shell (as was pointed out in the comments). The moment of inertia of a spherical shell is quite ...


Whether the upward direction will be taken as positive & the downward direction as negative or vice versa simply depends upon you. It does not bother any physics. Generally, the downward direction is taken as negative. The negative sign indicates the opposite direction only of what you have assigned the positive direction. In Principles of Physics by ...


It depends on what direction you assign to be positive in your coordinate system. To avoid confusion, just remember which direction acceleration is acting and which direction you assigned to be positive.


Well, things don't fall up, so gravity would be negative. The longer version: For classical mechanics, the direction of a coordinate system is often arbitrary. For free fall problems I generally choose down to be positive (especially when I'm not dealing with an initial velocity, but it's a matter of preference). What you do need to know, no matter what ...


If up is positive, and gravity points down, then $a$ (acceleration due to gravity) would be downwards, so it will have a negative magnitude.


The point at rest in $k$ moves as $(x(t),y(t),z(t)) = (x_0+vt,y,z)$ in the frame in which $k$ itself moves with $v$ in positive $x$-direction. Thus, we have to set $x'(t)=x(t)-vt = x_0 +vt-vt$ to obtain a stationary $x'(t) = x_0$ coordinate for a tupel $(x',y,z)$.


It's just a drawing convention. Rather than "vertical", time is orthogonal to the x-axis. The reason it is not shown vertical is because the paper surface is 2D, and the author uses the vertical axis for drawing the altitude with respect to the ground. Just recall the way you draw the 3D axis. Here, the author uses X (horizontal),Y (vertical) and time ...

Top 50 recent answers are included