Physics Help Forum ball on ramp experiment

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 Feb 5th 2017, 03:04 PM #1 Junior Member   Join Date: Feb 2017 Posts: 14 ball on ramp experiment I am trying to find the speed of the ball going down the ramp. I have the following info: ball mass ramp length ramp angle ramp friction (and ball radius/diameter if needed but I would like to not have to use that) is it possible to figure out the speed with just that info and if so anyone know an equation(even if it consists of 1000 different steps/equations)? Last edited by Zman; Feb 5th 2017 at 03:17 PM.
 Feb 5th 2017, 04:00 PM #2 Senior Member     Join Date: Jun 2016 Location: England Posts: 1,006 The basic equation you require is F=ma or force equals mass times acceleration Turn it around to get a=F/m You know the mass so you need to work out the force. The force is the force due to gravity, modified by the fact that you are going down a ramp, so you use trigonometry (sine of the angle) to determine the reduced effect of gravity. Next you have the Force due friction, which acts in the opposite sense to the force due to gravity. F(actual)=F(gravity)*sine(angle)-F(friction) Finally there is v=u+at or Final velocity is initial velocity plus the acceleration multiplied by the time the acceleration has been acting. (if you are starting at zero velocity then this becomes just v=at) Note that the acceleration due to gravity (without a ramp) is 9.82 meters per second per second. topsquark likes this.
 Feb 5th 2017, 05:41 PM #3 Junior Member   Join Date: Feb 2017 Posts: 14 I'm a bit confused on paragraphs 3 and 4 the paragraphs about force of gravity and friction can you go into more detail please
Feb 5th 2017, 06:07 PM   #4

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 Originally Posted by Zman I'm a bit confused on paragraphs 3 and 4 the paragraphs about force of gravity and friction can you go into more detail please
Gravity is a force between two masses: In this case we are talking about the weight of the ball. This force always points directly to the floor, and has a magnitude of w = mg. Since the mass is moving down the incline we need to break the force into components along the incline and perpendicular to the surface. Make a sketch of it and use some basic trigonometry.

Friction is a retarding force that points in the direction opposite to the direction of motion. There are two types: static and kinetic. From the problem statement we are taking the motion to be down the incline, so friction will act up the incline with a force of $\displaystyle f_k = \mu _k N$ where $\displaystyle \mu _k$ is the coefficient of kinetic friction and N is the normal force from the incline.

-Dan
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 Feb 6th 2017, 12:34 AM #5 Physics Team   Join Date: Feb 2009 Posts: 1,425 When the ball rolls down the slope, it has rotational kinetic energy also. Thus the equation would be mgh = 1/2 mv^2 + 1/2 I (omega)^2 where I is the moment of inertia and omega the angular velocity. I for a solid sphere is I = 2/5 m r^2 and 2/3 m r^2 for a thin spherical shell. Since v = r w (w = omega) giving mgh = m (7/10) v^2 from which you can work out the final velocity if that is what you are looking for. When a ball or wheel rolls, the point in contact does not move, so the friction is static, and this is what lets the ball roll. The centre will move at v, and the point diametrically opposite the point in contact will move at 2v. Friction in rolling does not contribute as much to slowing down as in the case of sliding, so the net effect will be less than μ mg cos theta.
 Feb 6th 2017, 08:10 AM #6 Junior Member   Join Date: Feb 2017 Posts: 14 You're right. I forgot the rotation of the ball. Can you make that simpler or use less confusing symbols?(I'm a noob)
 Feb 6th 2017, 09:55 AM #7 Physics Team   Join Date: Feb 2009 Posts: 1,425 The symbols used are more or less the standard ones. The equation refers to conservation of energy. mgh is the initial potential energy, When it rolls down, we may write m g h= 1/2 m v^2 + 1/2 I ω^2-work done against friction, where the first term on the right is the K.E. and the second the rotational K.E. I stands for the moment of inertia, and ω is the angular velocity, which is the angle ( in radians )thru' which it rotates per sec If the symbol ^ is confusing, it means raising something to a power. Could you clarify what exactly is not clear
 Feb 6th 2017, 11:28 AM #8 Junior Member   Join Date: Feb 2017 Posts: 14 That all clears it up although what does the omega stand for? Last edited by Zman; Feb 6th 2017 at 11:33 AM.
 Feb 7th 2017, 07:00 AM #9 Senior Member   Join Date: Aug 2010 Posts: 434 ?? Physicsquest said "$\displaystyle \omega$ is the angular velocity". Do you know what that is? It is the rate at which the ball is rotating, in "degrees per second" or "radians per second" or, possibly, "rotations per second" depending upon how you measure the angles.
 Feb 7th 2017, 07:01 AM #10 Senior Member   Join Date: Aug 2010 Posts: 434 Accidental double post. Last edited by HallsofIvy; Feb 7th 2017 at 07:03 AM.

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