# Rolling a ball off a roof (ramp)

• physicsfailure
In summary, Haruspex is trying to solve a homework problem involving a ramp and a ball. He is unsure of how to start and needs help.
physicsfailure

## Homework Statement

In this problem, I'm rolling a ball off a ramp. The ramp is supposed to simulate a roof. The hypotenuse of the ramp is 1.5 meters and the angle is 45 degrees. The distance from the edge of the ramp to the ground is -5.43 meters. That is the information I'm given. From there, I'm supposed to find the following:
G = -10
V = ?
Ramp Angle = 45°
Y = ?
Vyi = ?
Vxi = ?
Fall Time = ?
D = ?
Ramp Height = ?

## Homework Equations

SOH, CAH, TOA
V = √2GY
D = Vit + 1/2gt^2
T = √(2d/a)

## The Attempt at a Solution

To find ramp height:
1.5sin(45) = 1 meter.
To find V:
√2*-5.43*-9.8 = 10.3m/s
After that, I wasn't sure what to do.
My thought was that maybe to solve for Vxi and Vyi, I would use SOH, CAH, TOA?
Therefore:
10.3cos(45) = 7.2
10.3sin(45) = 7.2
??
I am really unsure about how to approach this problem, any help would be appreciated. Thanks

To find V:
√2*-5.43*-9.8 = 10.3m/s

That would be the vertical velocity when it hits the ground if simply dropped from 5.43 meters.

However I understand it the ball rolls down the roof picking up speed and then falls 5.43 meters. So in the equation..

V2=U2+2as

..the initial velocity U isn't zero.

physicsfailure said:
V = ?
Y = ?
Vyi = ?
Vxi = ?
D = ?

SOH, CAH, TOA

V = √2GY
D = Vit + 1/2gt^2
Is that the same D as above?
1.5sin(45) = 1 meter.
Somewhat inaccurate.
To find V:
√2*-5.43*-9.8 = 10.3m/s
I don't understand how you get that equation.

Haruspex: I am not trying to make people "guess" my variables. My teacher is very unclear and I have a very hard time understanding the subject matter when he explains it and I unfortunately had to miss some class time which didn't help either. According to the formula sheet, he has it noted that y is the distance in the y direction (in which case, I suppose that is also given with the -5.43m) and D is just distance. I assume D would be representative of how far out the ball lands once it leaves the ramp.

Cwatters: Would initial velocity be when the ball leaves the ramp or first starts rolling down the ramp? Sorry if this is a stupid question but I'm very new to physics.

physicsfailure said:
: Would initial velocity be when the ball leaves the ramp or first starts rolling down the ramp? .
Initial and final velocities are whatever you define them to mean. There are two phases of movement here, the rolling down the roof and the fall from there to the ground. Each phase has an initial velocity in each direction (X and Y), and a final velocity in each. The final velocities for the first phase are, of course, the initial belocities for the second phase. It helps to define separate variables for each of the six quantities.
There is a small difficulty with the problem as stated. A roling ball has some of its energy in rotational motion, but we're not given the ball's radius so we cannot calculate that. In effect, we have to treat it instead as if it slides down a frictionless ramp.
So, what are the velocities when the ball reaches the edge of the ramp? What equation can you use to find how long it will take to reach the ground from there?

physicsfailure said:
Cwatters: Would initial velocity be when the ball leaves the ramp or first starts rolling down the ramp?

As haruspex said this is a problem in two stages..

1) Rolling down the roof.
2) Falling from the edge of the roof to the ground.

You appear to have attempted stage 2) first but have assumed the initial velocity for stage 2) is zero.

Start again. You need to solve stage 1) first. When you know the final velocity at the end of stage 1) use that as the initial velocity for stage 2).

haruspex:
For time, I could manipulate these two formulas: Vyf = vyi + gt and Y = Vyit + 1/2gt^2. My teacher says to use the first formula but you must first solve for Vyf.
However, I'm finding it difficult to find a formula for velocity where I don't need Vi of Vf. I guess my question would be, in the first phase, where the ball is being rolled down the ramp, would Vi be 0? In which case I would get that same 10.3 m/s for Vf (or Vi in the second phase?) as I got in my original post by using the Vf^2 = Vi^2 + 2gy unless I made a mathematical error.
EDIT:
Actually, the number I think I should be getting is -4.4m/s. For y in the formula (distance traveled in y direction) I believe I should be using -1. I got this number by using 1.5sin(45) to get me the height of the triangle.

Last edited:
physicsfailure said:
haruspex:
For time, I could manipulate these two formulas: Vyf = vyi + gt and Y = Vyit + 1/2gt^2. My teacher says to use the first formula but you must first solve for Vyf.
However, I'm finding it difficult to find a formula for velocity where I don't need Vi of Vf. I guess my question would be, in the first phase, where the ball is being rolled down the ramp, would Vi be 0? In which case I would get that same 10.3 m/s for Vf (or Vi in the second phase?) as I got in my original post by using the Vf^2 = Vi^2 + 2gy unless I made a mathematical error.
EDIT:
Actually, the number I think I should be getting is -4.4m/s. For y in the formula (distance traveled in y direction) I believe I should be using -1. I got this number by using 1.5sin(45) to get me the height of the triangle.
As I said, 1.5 sin(45) = 1 is a bit inaccurate.
Let's deal with the first phase first. The ball is not falling freely, so its vertical acceleration is not g. There are two approaches we can use here. Either consider the forces on the ball as it rolls down and thus find the vertical component of acceleration, or use conservation of work to find the speed of the ball when it leaves the roof.

haruspex said:
As I said, 1.5 sin(45) = 1 is a bit inaccurate.

Why is it inaccurate? What should I do if not that?

haruspex said:
Either consider the forces on the ball as it rolls down and thus find the vertical component of acceleration

How do I do this?

physicsfailure said:
Why is it inaccurate? What should I do if not that?
What is sin(45)? It isn't 2/3.
How do I do this?
Draw a free body diagram. What are the forces on the ball? What is the net force parallel to the slope? What is the net force perpendicular to the slope? What equations do you get?
Or, using energy, what is the kinetic energy of the ball as it leaves the roof?

Haruspex: I cannot use energy equations because I am not given the mass of the ball. Thanks for the help but I guess I am just not getting this, lol.

physicsfailure said:
Haruspex: I cannot use energy equations because I am not given the mass of the ball. Thanks for the help but I guess I am just not getting this, lol.

So let the mass of the ball be m. It will cancel out.

haruspex said:
So let the mass of the ball be m. It will cancel out.

1.5sin(45) = 1.06 (more accurate, used fully accurate number when doing calculations on the calculator)
PE = mgh
PE = m*-9.8*-1.06
10.3m = 7/10mv^2 (moment of inertia)
M's cancel out...
10.3 = 7/10v^2
10.3/ (7/10)
14.84 = v^2
√14.84 = 3.85227m/s = Vf

Is this correct? Should it be -3.857227?

physicsfailure said:
1.5sin(45) = 1.06 (more accurate, used fully accurate number when doing calculations on the calculator)
PE = mgh
PE = m*-9.8*-1.06
10.3m = 7/10mv^2 (moment of inertia)
M's cancel out...
10.3 = 7/10v^2
10.3/ (7/10)
14.84 = v^2
√14.84 = 3.85227m/s = Vf

Is this correct? Should it be -3.857227?
Yes, that all looks good. Remember, this is net speed. You still have to separate it into vertical and horizontal components, so the sign is kind of irrelevant as yet.
Well done for spotting that I was wrong to say you cannot take rolling into account.

haruspex said:
Yes, that all looks good. Remember, this is net speed. You still have to separate it into vertical and horizontal components, so the sign is kind of irrelevant as yet.
Well done for spotting that I was wrong to say you cannot take rolling into account.

How do I separate this into vertical and horizontal components? Would it be something along the lines of 3.8sin(45)? (I'm probably completely wrong that was just a guess)

physicsfailure said:
How do I separate this into vertical and horizontal components? Would it be something along the lines of 3.8sin(45)? (I'm probably completely wrong that was just a guess)

Good guess.

haruspex said:
Good guess.
3.852272sin(45) = 2.72396
3.852272cos(45) = 2.72396
These numbers are vxi and vyi (initial velocities in the x and y direction)?
Would I use the original 3.852272m/s to solve for V final in the following formula: Vf^2 = Vi^2 + 2gy?

physicsfailure said:
3.852272sin(45) = 2.72396
3.852272cos(45) = 2.72396
These numbers are vxi and vyi (initial velocities in the x and y direction)?
Would I use the original 3.852272m/s to solve for V final in the following formula: Vf^2 = Vi^2 + 2gy?
Sorry, no internet access for a few days.
Yes, depending on what you are putting for y.
do you also need to find the time to land and the horizontal displacement?

## 1. How does the height of the ramp affect the distance the ball travels?

The height of the ramp affects the potential energy of the ball as it rolls off the roof. The higher the ramp, the greater the potential energy, and therefore the greater the distance the ball will travel. This can be explained by the law of conservation of energy, where the potential energy is converted into kinetic energy as the ball rolls down the ramp.

## 2. What is the relationship between the angle of the ramp and the distance the ball travels?

The angle of the ramp also affects the distance the ball travels. The steeper the angle, the greater the potential energy and the longer the distance the ball will travel. This is because a steeper ramp allows the ball to gain more speed as it rolls down, resulting in a longer distance traveled.

## 3. How does the surface of the ramp impact the motion of the ball?

The surface of the ramp can have a significant impact on the motion of the ball. A rough surface will create more friction, slowing down the ball and resulting in a shorter distance traveled. A smooth surface, on the other hand, will reduce friction and allow the ball to roll further.

## 4. What factors affect the velocity of the ball as it rolls off the ramp?

The velocity of the ball is affected by multiple factors, including the height and angle of the ramp, the surface of the ramp, and the weight and shape of the ball. These factors all contribute to the potential energy and speed of the ball as it rolls off the ramp.

## 5. What is the role of gravity in rolling a ball off a roof?

Gravity plays a crucial role in the motion of the ball as it rolls off a ramp. Gravity is constantly pulling the ball downwards, and as the ball gains potential energy from the height of the ramp, gravity converts it into kinetic energy, causing the ball to roll down the ramp. Without the force of gravity, the ball would not roll off the ramp at all.

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