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

• GrimesA
In summary, the conversation discusses a physics problem involving rotational energy, momentum, trajectories, inertia, and inclined planes. The problem involves predicting the final distance a solid sphere will land after rolling down an inclined plane and off of a table, taking into account a 4% energy loss. The conversation also includes a discussion on how to solve the problem algebraically, with one person breaking it into two separate problems and another providing guidance on how to approach the trajectory problem. The conversation also addresses the need for information on the angle of the inclined plane and the height of the release point from the floor in order to solve the problem accurately. Finally, there is a suggestion to use a formula involving 5/7 of the total kinetic energy to determine the
GrimesA

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

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.

Last edited by a moderator:
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.

gneill said:
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.

GrimesA said:
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.

gneill said:
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.

GrimesA said:
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.

gneill said:
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) ?

GrimesA said:
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.

gneill said:
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.

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.

dean barry said:
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

GrimesA said:
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.

gneill said:
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

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

gneill said:
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

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?

gneill said:
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

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

gneill said:
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?

GrimesA said:
If I look at the rotational vs the translational it'll be half. Correct?
You're saying that A = B? Always?

gneill said:
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

GrimesA said:
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?

gneill said:
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)

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

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).

## 1. What is an inclined plane?

An inclined plane is a flat surface that is angled or sloped, typically used to make it easier to move objects from one height to another.

## 2. How is the distance a sphere travels on an inclined plane calculated?

The distance a sphere travels on an inclined plane can be calculated using the formula: d = (g * sinθ * t^2)/2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), θ is the angle of inclination, and t is the time.

## 3. Does the mass of the sphere affect the distance it travels on an inclined plane?

Yes, the mass of the sphere does affect the distance it travels on an inclined plane. According to Newton's Second Law of Motion, the force required to move an object is directly proportional to its mass. Therefore, a heavier sphere will require more force to move and will travel a shorter distance on an inclined plane.

## 4. How does the angle of inclination affect the distance a sphere travels on an inclined plane?

The angle of inclination has a direct impact on the distance a sphere travels on an inclined plane. The steeper the angle, the shorter the distance the sphere will travel. This is because a steeper angle of inclination creates a greater force of gravity, causing the sphere to accelerate faster and reach the bottom of the inclined plane in a shorter amount of time.

## 5. Are there any external factors that can affect the distance a sphere travels on an inclined plane?

Yes, there are external factors that can affect the distance a sphere travels on an inclined plane. These factors include air resistance, friction between the sphere and the inclined plane, and the surface of the inclined plane (smoothness, material, etc.). These factors can alter the acceleration and overall distance traveled by the sphere on the inclined plane.

• Introductory Physics Homework Help
Replies
10
Views
792
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
19
Views
3K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
16
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
34
Views
3K
• Introductory Physics Homework Help
Replies
14
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
3K