What's the distance a sphere travels from inclined plane?

[GrimesA]

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >[/color]

So the problem that I have been assigned has formulas of rotational energy, momentum, trajectories, inertia, and inclined planes. A solid sphere is rolling down an inclined plane (that is placed on a table) and then off of the table (losing 4% of energy before it leaves the table). I must predict where the final distance on the ground the ball will land and show algebraic solutions.

I broke the part into two separate problems. One of inclined plane/rotational energy which leads me to the trajectory problem.

To set up the inclined plane to solve for the velocity that it leaves the table at I found that PE=KE. Therefore mgh=½mv^2 + ½((2/5)MR^2)(v/R)^2 which I then simplified to 2g*h=v^2(1+(2/5)) and finally to solve for Vf I said √2g*h (minus the 4%)/(1+(2/5)) = Vf

Now it is time for the part I was confused on. I originally started the trajectory problem solving in the Y to find the time it'll hit the ground in. I found that Yf= Yi+Vit+½at^2 and so t=√2*Yi/gravity which I then used to find Xf by saying Xf=Vi*t.

However, upon inspection I realized that I hadn't accounted for the angle of the inclined plane nor the fact that when I went from Yf= Yi+Vit+½at^2 to being t=√2*Yi/gravity I had made Vi of Y to be zero which I know applies to situation that the object is DROPPED where this was is already moving in both X and Y as the problem started. I am cloudy-minded on how to rectify my incorrections.

 

[gneill]

You need to add some information. What is the angle of the inclined plane (or the length L of its base)? That will let you determine the initial launch angle for the sphere. You also need to know the height of the release point (table top) with respect to the floor. Then, along with your launch speed, you'll have a classic trajectory problem.

 

[GrimesA]

You need to add some information. What is the angle of the inclined plane (or the length L of its base)? That will let you determine the initial launch angle for the sphere. You also need to know the height of the release point (table top) with respect to the floor. Then, along with your launch speed, you'll have a classic trajectory problem.

The angle was approximately 9 so I thought the angle of the trajectory would be 81. And the length will vary because it won't necessarily be dropped from the top of the plane.

 

[gneill]

The angle was approximately 9 so I thought the angle of the trajectory would be 81. And the length will vary because it won't necessarily be dropped from the top of the plane.

Your leftmost image indicates that the edge of the ramp is at the edge of the table top, and so the sphere will launch at the bottom of the ramp. It will launch at the same angle as the ramp (only below the horizontal).

You still need to know the height of the table top from the floor.

 

[GrimesA]

Your leftmost image indicates that the edge of the ramp is at the edge of the table top, and so the sphere will launch at the bottom of the ramp. It will launch at the same angle as the ramp (only below the horizontal).

You still need to know the height of the table top from the floor.

What an elementary mistake I made.
The distance from the floor to the table was approx 140 cm but again I must find algebraic solutions.

 

[gneill]

What an elementary mistake I made.
The distance from the floor to the table was approx 140 cm but again I must find algebraic solutions.

Well, declare variable names for the angle and table height. Then proceed with the trajectory problem given the launch angle, speed, and initial height.

 

[GrimesA]

Well, declare variable names for the angle and table height. Then proceed with the trajectory problem given the launch angle, speed, and initial height.

My dilemma was that I was unsure what the initial y velocity is because it IS in fact moving in the y direction. Would it be Vf(incline)*sin(Height from table to ball/length of table to ball) ?

 

[gneill]

My dilemma was that I was unsure what the initial y velocity is because it IS in fact moving in the y direction. Would it be Vf(incline)*sin(Height from table to ball/length of table to ball) ?

I suppose that will do. Presumably your ramp angle is constant so any way you determine it is fine.

You'll want to be careful with the sign you assign to the vertical velocity in order to be consistent with your choice of coordinate system for the trajectory portion of the problem.

 

[GrimesA]

I suppose that will do. Presumably your ramp angle is constant so any way you determine it is fine.

You'll want to be careful with the sign you assign to the vertical velocity in order to be consistent with your choice of coordinate system for the trajectory portion of the problem.

Sweet! I greatly appreciate the guidance! I think I know where to go from your advice.

 

[dean barry]

I would do this for part 1.
m*g*h * 0.96 = Total KE at leaving incline.
5/7 of this KE is attributable to linear velocity.
(typical for non-slipping homogenous sphere)
Then use: sqrt ( linear KE / ( ½ * mass ) ) to get velocity at leaving incline.
Split into horizontal and vertical vectors.

 

[GrimesA]

I would do this for part 1.
m*g*h * 0.96 = Total KE at leaving incline.
5/7 of this KE is attributable to linear velocity.
(typical for non-slipping homogenous sphere)
Then use: sqrt ( linear KE / ( ½ * mass ) ) to get velocity at leaving incline.
Split into horizontal and vertical vectors.

I'm lost on why the 5/7 is there. All I know is 2/5*m*r^2 is the energy used to rotate

 

[gneill]

I'm lost on why the 5/7 is there. All I know is 2/5*m*r^2 is the energy used to rotate

Write a general expression for the total kinetic energy (Translational and Rotational) of the rolling sphere and find the fraction of the total that "belongs to" translational motion.

 

[GrimesA]

Write a general expression for the total kinetic energy (Translational and Rotational) of the rolling sphere and find the fraction of the total that "belongs to" translational motion.

That's my 2mgh*.96=mv^2 + (2/5)mr^2 * (v/r)^2

From there only mv^2 is the translational

 

[gneill]

No, forget the current problem for a moment and write an expression for the total kinetic energy of a rolling sphere.

 

[GrimesA]

No, forget the current problem for a moment and write an expression for the total kinetic energy of a rolling sphere.

(.5)mv^2 + +(.5)*inertia*rotational velocity

 

[gneill]

Right. I apologize, you could actually have used your expression in your post above. I saw the 2mgh... on the left hand side and posted before realizing it wouldn't matter to what follows.

So, taking your expression for (twice) the total KE:

2KE = mv^2 + (2/5)mr^2 * (v/r)^2

What fraction of that total is translational KE?

 

[GrimesA]

Right. I apologize, you could actually have used your expression in your post above. I saw the 2mgh... on the left hand side and posted before realizing it wouldn't matter to what follows.

So, taking your expression for (twice) the total KE:

2KE = mv^2 + (2/5)mr^2 * (v/r)^2

What fraction of that total is translational KE?

That I am completely clueless on

 

[gneill]

You have a sum of two quantities. Say A + B. What is an expression for the fraction of A + B that A represents?

 

[GrimesA]

You have a sum of two quantities. Say A + B. What is an expression for the fraction of A + B that A represents?

If I look at the rotational vs the translational it'll be half. Correct?

 

[gneill]

If I look at the rotational vs the translational it'll be half. Correct?

You're saying that A = B? Always?

 

[GrimesA]

You're saying that A = B? Always?

That's why I was confused because they won't always be the same. If the radius of the sphere is greater then it'll have more rotational, if velocity is greater then translational will be greater

 

[gneill]

That's why I was confused because they won't always be the same. If the radius of the sphere is greater then it'll have more rotational, if velocity is greater then translational will be greater

Sure. So let's break it down to a simple example. I give you two numbers, 30 and 70. Clearly they sum to 100. What fraction of the total 100 does the 30 represent? How would you write this fraction using the given numbers?

 

[GrimesA]

Sure. So let's break it down to a simple example. I give you two numbers, 30 and 70. Clearly they sum to 100. What fraction of the total 100 does the 30 represent? How would you write this fraction using the given numbers?

In that case it'll be A/(A+B)

 

[gneill]

Right. So, pick out the A and B terms from the total KE formula, write the fraction, and simplify.

 

[dean barry]

2/5 * m * r ² calculates the mass moment of inertia (i) of a solid homogenous sphere.

A (homogenous) solid sphere rolling without slipping will always have the same KE distribution ratio between linear and rotating.

You can check this by imagining the sphere at a constant velocity (v) of say 10 m/s (arbitrary)
Calculate first the linear KE from
linear KE = ½ * m * v ²
And the rotating KE from:
rotating KE = ½ * i * ω ²
(ω is the rotation rate in radians / second, get this from: ω = v / r)
(r is the sphere radius in metres)
The total KE = linear + rotating
The linear KE will be 5/7 of the total KE
The rotating KE will be 2/7 of the total KE


If the sphere rolls without slipping from the top of the incline then the PE (m*g*h) will be translated into KE, part of which will be linear KE and part rotating KE, but the distribution will be as described above (5/7 : 2/7).

So, calculate the PE translated (in Joules) from m*g*h
(this gets converted into linear KE and rotating KE as it rolls down the ramp, in the ratio described)
(m = mass in kg, g = local gravity rate (9.81 is OK) in (m/s)/s, h = vertical height in metres fallen by sphere while rolling down the incline.)
Find the portion of KE belonging to linear (5/7 * m*g*h)
To find the linear speed this relates to, use the equation :
linear KE = ½ * m * v ²
you have the linear KE, so translate the equation for velocity v:
v = square root ( linear KE / ( ½ * m ) )
This gives you the linear velocity at the bottom of the incline.
You can forget everything else now and work on the flight path now you have the velocity (and presumably the incline angle).