0
$\begingroup$

I understand that you resolve the components of the balls weight parallel and perpendicular to the slope in order to calculate the force due to gravity that actually acts down the slope.

The component parallel to the slope would be: $m g \sin(x)$ Normal to the slope would be: $mg\cos(x)$

cancelling the masses you get the acceleration down the slope to be $g\sin(x)$.

Can you then resolve this acceleration into its horizontal and vertical components? Would you also need to resolve the component normal to the slope once more into horizontal and vertical components if you can do so, and then add them?

Doing the above ( and leaving out the component normal to the slope) I got the vertical and horizontal acceleration to be:

vertical: $g\cos^2(x)$ horizontal: $g\cos(x)\sin(x)$

I want to be able to work out the horizontal and vertical components of the velocity of the ball once it has reached the bottom of the slope, but am not quite sure.

Any help would be greatly appreciated!

$\endgroup$
  • $\begingroup$ If $x$ is the angle between the slope and the horizontal, then the vertical acceleration is simply $g$ and the horizontal acceleration is $g \sin(x) \cos(x)$. Then work out how long the ball will be travelling in time $t$ before it reaches the bottom. You need the height of the slope for that. From that time calculate the vertical and horizontal velocity components. All this assumes the ball has no inertial moment or slides frictionlessly down the slope. $\endgroup$ – Gert Aug 24 '15 at 20:17
1
$\begingroup$

If velocity at bottom of slope is all you want then energy method in Gert's and Dr Xorille's answers is an elegant and easy way to get at it. However you seem to be tangled up with resolving vectors.

In working with vectors you should set up a coordinate system and stick to it until the end. If you have to set up more than one, and worse, have to switch between them often (as it is in your case), then you make sure that every component of a vector (in your case gravity vector $g$) is expressed in the same coordinate system.

In your example there are two coordinate systems: X-Y system in which X direction is parallel to ground and Y direction is perpendicular to ground and downward; P-Q system in which P direction is parallel to slope and Q direction is perpendicular to slope.

In X-Y system gravity vector $\textbf{g}\equiv(0,g)$ while in P-Q system $\textbf{g}\equiv(g\sin \theta,g\cos\theta)$. You may now solve the problem in any of the two coordinate systems of your choice.

What you have done is, you have calculated $\textbf{g}$'s components parallel and perpendicular to slope (i.e. in P-Q system) but for some reason you want to go back to X-Y system. That is fine but then you $\textit{must}$ take both components of $\textbf{g}$ in P-Q system and re-express them in X-Y system. Therefore you should not leave out the component normal to the slope, because both components, parallel and perpendicular to slope, together make the gravity vector $\textbf{g}$. Although doing this will simply return you to components of $\textbf{g}$ in X-Y system viz. $(0,g)$ and I don't see the point of it all; if you wanted to stay in X-Y system you could have done so right from the beginning.

Once you know components of $\textbf{g}$ in X-Y system you may now calculate velocity at any other instant by making use of the usual laws of motion.

$\endgroup$
0
$\begingroup$

Another way to do this is to consider the energy (as you do not seem to be losing energy anywhere, eg with friction).

You have an initial velocity (perhaps 0) which corresponds to a kinetic energy, and a potential energy (which is $mgh$). Then the potential energy is converted to kinetic energy, and you can simply figure it out from there.

One important point, which @Gert also hinted at, is that rolling is very different from sliding. In rolling, you have both kinetic energy associated with average velocity, and kinetic energy associated with the rotation. The rotational energy requires the moment of inertia. If the ball is rolling without slipping, then the rotational energy will be related to the velocity.

$\endgroup$
  • $\begingroup$ Dr Xorile: using conversion of potential energy to kinetic energy is of course by far the easiest way to calculate final velocity (assuming sliding), then calculate the hor. and vert. components of $v$. But it seems the OP wants to do it the 'long way', by means of component accelerations. $\endgroup$ – Gert Aug 24 '15 at 22:55
  • $\begingroup$ Possibly. He wants to figure out the vertical and horizontal components, and has tried to resolve it. I thought it at least worth mentioning. ;-) $\endgroup$ – Dr Xorile Aug 25 '15 at 3:15
0
$\begingroup$

Assume the ball slides down the slope without friction and that it starts from stationary ($v=0$) at a height above the horizontal $h$.

During the friction free slide the object's potential energy $mgh$ is converted to kinetic energy $\frac{mv^2}{2}$, so that:

$mgh=\frac{mv^2}{2}$ and:

$v=\sqrt{2mgh}$.

This speed vector is of course oriented parallel to the slope. The horizontal component of $v$, say $v_x$, is:

$v_x=\cos(x) v$.

While the vertical component, say $v_y$, is:

$v_y=\sin(x) v$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.