# Rotational Motion/Projectile motion of a ball on a circular ramp.

• andorrak
In summary, the sphere rolls without slipping on the track shown in the figure whose radius is R0 = 40.0 cm. The sphere starts rolling at a height R0 above the bottom of the track. It leaves the track after passing through an angle of 135°.
andorrak

## Homework Statement

A small sphere of radius r0 = 1.4 cm rolls without slipping on the track shown in the figure whose radius is R0 = 40.0 cm. The sphere starts rolling at a height R0 above the bottom of the track. Assume that it leaves the track after passing through an angle of 135° as shown.

Picture: http://imgur.com/7YLlJ

It asks for speed at the end of the ramp. and what is d.

## Homework Equations

I know all the basic equations. I am just confused how to get the height at the end of the ramp. ie. here's my equation.

E0=mgh or m(9.8)(.4)

E1=.5mv2+.5Iw2 +mgh or

I replace "I" with the moment of inertia for a perfect sphere which is 2/5mr2

.5mv2+.5(2/5mr2)v2+m(9.8)(h)

E1 is the energy at the end of the ramp. But i cannot get h which is that height at the end of the ramp.

I figure the angle of that circle must be 45 degrees. So that arc length is .31m. Since the Radius is 40cm. Arc Length = angle times radius.

But i cannot get the height. If i get the height i can solve for the speed, then the get d.

Try to draw the unknown height into the picture before you say 'I can not get h'. See attachment.

ehild

#### Attachments

• ramp.JPG
7.8 KB · Views: 1,089
how can you say that at the point of the circle will make a full radius above that point?

Yea you absolutely cant. the triangle would have two lengths of radius 40 cm repeated on the hypotenuse and the sides.

andorrak said:
how can you say that at the point of the circle will make a full radius above that point?

I certainly have not said that.

ehild

then i have no idea what is going on geometrically

Examine the triangle. What is the length of the hypotenuse?
What is its angle?

ehild

haha i feel like an idiot now. math does that to you.

but how do i get d. i think my kinematics get me the distance from the point it shot off the ramp. i can't make a triangle underneath to get it.

But my answers don't seem to be right. here was my math.

I had h being .1171 m.

so the velocity turned out to be 1.76 m/s which is wrong.

I redid the math as I was looking at my work and i don't know what is wrong just wanted to check that with you.

I get:

9.8*.4=.5v2+(1.4^2)(.5)(.4)v2+9.8*.1171

2.77242=.892v2

v=1.76

andorrak said:
I get:

9.8*.4=.5v2+(1.4^2)(.5)(.4)v2+9.8*.1171

You count the translational KE twice.
Edit: and included Ro into the rotational energy, why?

ehild

Last edited:

flyingpig said:

The centripetal force does not "act" it is just the resultant of other forces. You might have thought of the normal force. But it is normal to the trajectory, so its work is zero.

ehild

Is it possible to solve using torque and other Newton's laws?

Feel abd for hijacking (I won't show the steps), but I got roughly 1.98m/s

flyingpig said:
Is it possible to solve using torque and other Newton's laws?

Feel abd for hijacking (I won't show the steps), but I got roughly 1.98m/s

You can solve it by other methods than conservation of energy, but it seems the simplest. Your result is about correct.

ehild

You count the translational KE twice.
Edit: and included Ro into the rotational energy, why?

ehild

I thought I am supposed to count both the kinetic energy of the ball and the rotational energy, which I thought accounts for the energy total. and what is Ro?

andorrak said:
I thought I am supposed to count both the kinetic energy of the ball and the rotational energy, which I thought accounts for the energy total. and what is Ro?
Yes, but what is the rotational energy? Ro is the radius of the track in the figure, 0.4 m.

ehild

I thought the rotational energy is the one that involves the sphere. The 1.4cm

So 1.4 was cm and the radius of the sphere! Now I understand where is it from, and what is 0.4. But: The rotational energy is not 0.5 I v^2, but 0.5 I ω^2 (ω is the angular velocity).

ehild

I use 0.4 m in the beginning of the equation to represent the PE of the initial state before it begins to roll. I know that the rolling energy is .5Iw^2. But i replace I with 2/5r^2 which is the moment of inertia for a sphere. so the .4 here might be the confusion u are talking about.

hence I get for the rolling energy of the sphere:

.5*.4*(1.4/100)^2 *v^2

I also realized I've been using 1.4 METERS as the radius. Derp. but i still do not get the answer. With the 1.4 cm use I get the velocity as 2.35 m/s. Which i know is wrong. Any suggestions Ehild?

andorrak said:
I know that the rolling energy is .5Iw^2. But i replace I with 2/5r^2 which is the moment of inertia for a sphere. so the .4 here might be the confusion u are talking about.

hence I get for the rolling energy of the sphere:

.5*.4*(1.4/100)^2 *v^2

You can not use v instead of w. v is the velocity of the CM, w is the angular speed of rotation. How are they related in case of pure rolling?
Check the dimension of your formula. Disregarding mass, it has the unit of m^4/s^2 instead of m^2/s^2.

ehild

Of course i know that dude.

but remember Omega = V/R? so the rotational energy is .5IW^2. I was simply canceling out R ahead of time.

I= .4R^2 and we have (V/R)^2. the R's cancel out So i just get energy.

Just went and talked with my professor and he showed me what i did wrong:

I had everything in place correctly EXCEPT i did not cancel out the r's as I had stated before, ie:

Left side: 9.8*.4

Right Side: .5v^2+.5*.4*(R^2)*(V^2/R^2) + (9.8)(.1171)

My v then becomes 1.99m/s.

The r^2's cancel out with the bottom denominator of the omega term.

Thanks for all your help ehild.

wait but how do i get the distance underneath the ramp?

Like i got the distance from the moment it leaves the ramp to the end but not before.

Last edited:

## 1. What is rotational motion?

Rotational motion is the movement of an object around a fixed axis. This type of motion is characterized by the object's rotation or spin.

## 2. How does a ball move on a circular ramp?

A ball on a circular ramp will experience both rotational and projectile motion. As it moves around the ramp, it will also be propelled upward, creating a curved path.

## 3. What factors affect the rotational motion of a ball on a circular ramp?

The rotational motion of a ball on a circular ramp can be affected by the ball's mass, the radius of the ramp, and the force applied to the ball.

## 4. How is the velocity of a ball on a circular ramp calculated?

The velocity of a ball on a circular ramp is calculated using the formula v = ωr, where v is the linear velocity, ω is the angular velocity, and r is the radius of the ramp.

## 5. What is the difference between rotational and projectile motion?

Rotational motion involves the movement of an object around a fixed axis, while projectile motion involves the motion of an object through the air under the influence of gravity. A ball on a circular ramp experiences both types of motion simultaneously.

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