Frictionless half-pipe and rotational motion question

In summary, the conversation discusses a problem involving a block sliding down a frictionless half-pipe and determining its velocity, tangential acceleration, radial acceleration, and angular acceleration at various angular displacements. The conversation covers the use of rotational motion equations, converting degrees to radians, using energy conservation, and finding the height as a function of angle in order to solve for the tangential velocity. Further discussion includes the use of trigonometry to solve for the angular acceleration and the formula for tangential acceleration. The conversation concludes with a question about attending McMaster University.
  • #1
AJDangles
48
0

Homework Statement



A block slides down a frictionless half-pipe (shown on the back of this page). It is released from rest at the point (2,0).

Determine the velocity, tangential accleration, radial acceleration and angular acceleration when the block goes through angular displacements of 30, 60, 90, 120, 150 and 180 degrees respectively.

On the diagram, draw vectors to scale representing each of the linear quantities.

Homework Equations



θ = s/r

Really all of the rotational motion equations. Including the rotational kinematics equations.

The Attempt at a Solution



I calculated the displacement at each degree interval by converting the degress to radians and then using the equation θ = s/r.

My question to you is where do I go from here? The question doesn't give me anything else other than that it's dropped from rest.

Am I to assume that the radial acceleration is equal to gravity? Any help would be much appreciated.

Image of half-pipe attached.
 

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  • #2
AJDangles said:
Am I to assume that the radial acceleration is equal to gravity?
Why assume that? What do you need to calculate the radial acceleration? (What does it depend on?)
 
  • #3
Doc Al said:
Why assume that? What do you need to calculate the radial acceleration? (What does it depend on?)

It depends on the tangential velocity and the radius, right?

a = v^2/r

Or angular velocity and radius which is:

a = (w^2)r

I guess my question is: How can I approach this problem to determine the tangential velocity or the angular velocity? Does that sound about right?
 
Last edited:
  • #4
AJDangles said:
It depends on the tangential velocity and the radius, right?

a = v^2/r

Or angular velocity and radius which is:

v = (w^2)r
Good!
I guess my question is: How can I approach this problem to determine the tangential velocity or the angular velocity? Does that sound about right?
You're on the right track. Now how can you figure out the speed of the block after it slides down to some point on the path? (Hint: The pipe is frictionless.)
 
  • #5
Doc Al said:
Good!

You're on the right track. Now how can you figure out the speed of the block after it slides down to some point on the path? (Hint: The pipe is frictionless.)

Do I use Vf^2 = Vi^2 + 2ad

Where Vi = 0, Vf = ?, a = 9.8, and d = the distance in radians between each interval?
 
  • #6
AJDangles said:
Do I use Vf^2 = Vi^2 + 2ad

Where Vi = 0, Vf = ?, a = 9.8, and d = the distance in radians between each interval?
That won't work, at least not directly. That equation applies for constant acceleration only. (Such as something in free fall.)

Another hint: What's conserved?
 
  • #7
Doc Al said:
That won't work, at least not directly. That equation applies for constant acceleration only. (Such as something in free fall.)

Another hint: What's conserved?

Energy is conserved... but how do I calculate the kinetic energy without the mass?
 
  • #8
AJDangles said:
Energy is conserved... but how do I calculate the kinetic energy without the mass?
Just call the mass 'm' and proceed. Perhaps you won't need the actual mass.
 
  • #9
Doc Al said:
Just call the mass 'm' and proceed. Perhaps you won't need the actual mass.

Alright. So I have:

(0.5)m(Vi^2) + mgh = (0.5)m(Vf^2) + mgh
0 + (9.8*2) = 0.5(Vf^2) + 9.8(h)

For that last h in that second line of the equation; how do i determine that value? Should I simply eyeball it from the graph?
 
  • #10
AJDangles said:
For that last h in that second line of the equation; how do i determine that value? Should I simply eyeball it from the graph?
You have a half circle of given radius. You should be able to find the height as a function of angle. (Without eyeballing anything.)
 
  • #11
Ok. So I take sin30(2) = 1 for my h.

0 + (9.8*2) = 0.5(Vf^2) + 9.8(h)
which gives:
Vf = 2

And that is the tangential velocity.

Thank you very much for your help!
 
  • #12
AJDangles said:
Ok. So I take sin30(2) = 1 for my h.

Hey I am just wondering why you used sin here? I am sure the answer is simple but i just don't see it.
 
  • #13
AnanthanK said:
Hey I am just wondering why you used sin here? I am sure the answer is simple but i just don't see it.

Draw yourself a line from the origin to the circle at an angle of 30 degrees. If you draw a line from the x-axis to the point where your line from the origin meets the edge of the circle, you will have drawn a right triangle. The y component (i.e the height) is what you are looking for. Using simple trig to solve for the length of the y-component of the right angle triangle.
 
  • #14
AJDangles said:
Ok. So I take sin30(2) = 1 for my h.

0 + (9.8*2) = 0.5(Vf^2) + 9.8(h)
which gives:
Vf = 2

And that is the tangential velocity.

Thank you very much for your help!

Also, my math in this post is wrong, but it's the correct process
 
  • #15
Unrelated to Ananthan's question. Could someone else help me to find either the angular accleration or the tangential acceleration of the block? I'm stuck. I'm thinking it's something along the lines of determing the angular velocities using V = rw since we have now calculated the tangential velocity. Then you can take these angular velocities at each point and substitute into the equation

wf^2 = (wi^2) + 2(ang. accn)theta

to solve for angular acccleration.

Once you have the angular acceleration at each point... you can then then determine the tangential acceleration by use of the formula A = (ang accn)(radius)
 
  • #16
AJDangles said:
Draw yourself a line from the origin to the circle at an angle of 30 degrees. If you draw a line from the x-axis to the point where your line from the origin meets the edge of the circle, you will have drawn a right triangle. The y component (i.e the height) is what you are looking for. Using simple trig to solve for the length of the y-component of the right angle triangle.

Thank you very much!
 
  • #17
AnanthanK said:
Thank you very much!

No problem. Do you go to McMaster?
 
  • #18
Apparently i didn't search well enough before I posted. Thanks for responding to my post, which I'll now delete. And yes, I do go to Mac!
 
  • #19
AJDangles said:
Ok. So I take sin30(2) = 1 for my h.
Careful here. In general, sinθ(R) would give you the distance from the top of the pipe. (It doesn't matter for θ = 30°, but for other angles it will.)

0 + (9.8*2) = 0.5(Vf^2) + 9.8(h)
which gives:
Vf = 2

And that is the tangential velocity.
As you have already discovered, your arithmetic is wrong.
 
  • #20
AJDangles said:
Unrelated to Ananthan's question. Could someone else help me to find either the angular accleration or the tangential acceleration of the block? I'm stuck.
Use Newton's 2nd law.
 
  • #21
Doc Al said:
Use Newton's 2nd law.

OK. How do I go about doing this? I attended my lecture today and discovered that you needed to draw out the free-body diagram for the object at a specific point. The normal points towards the middle of the circle, gravity downwards and the tangential acceleration is along the edge of the circle. I'm thinking that the equation you would use would be something like this:

ma = (mass)(gravity)sin(theta) --> Just like you would do for an inclined plane.

Cancel out the masses... solve for acceleration? However, I'm thinking that the above equation there is incorrect, because taking sin would give me the horizontal component of the acceleration, would it not? If that isn't the case, then I'm probably visualizing my FBD completely incorrectly.
 
  • #22
AJDangles said:
OK. How do I go about doing this? I attended my lecture today and discovered that you needed to draw out the free-body diagram for the object at a specific point. The normal points towards the middle of the circle, gravity downwards and the tangential acceleration is along the edge of the circle.
All good. You need the force components tangent to the circle.

I'm thinking that the equation you would use would be something like this:

ma = (mass)(gravity)sin(theta) --> Just like you would do for an inclined plane.

Cancel out the masses... solve for acceleration? However, I'm thinking that the above equation there is incorrect, because taking sin would give me the horizontal component of the acceleration, would it not? If that isn't the case, then I'm probably visualizing my FBD completely incorrectly.
With the inclined plane you're given the angle with the horizontal, so the component of the weight parallel to the incline is mgsinθ. But here you need to find the angle that the tangent to the circle makes and relate that to the given angle θ. Draw yourself a careful diagram.
 
  • #23
Doc Al said:
All good. You need the force components tangent to the circle.


With the inclined plane you're given the angle with the horizontal, so the component of the weight parallel to the incline is mgsinθ. But here you need to find the angle that the tangent to the circle makes and relate that to the given angle θ. Draw yourself a careful diagram.

Like this? (See diagram)
 

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  • #24
AJDangles said:
Like this? (See diagram)
The diagram looks OK, but your calculation does not. What you've labeled as G, is the weight of the object, mg, which acts down. What's the component of that force in the tangential direction? (You have the angle correctly shown.)
 
  • #25
Doc Al said:
The diagram looks OK, but your calculation does not. What you've labeled as G, is the weight of the object, mg, which acts down. What's the component of that force in the tangential direction? (You have the angle correctly shown.)

cos30ma? Because the hypotenuse of that triangle is equal to the force in the tangential direction?
 
  • #26
AJDangles said:
cos30ma?
Nope.
Because the hypotenuse of that triangle is equal to the force in the tangential direction?
You are finding one component of the weight. When finding the component of a vector, the full vector is the hypotenuse. That means mg is the hypotenuse, and the tangential component is one of the sides of the right triangle.
 
  • #27
Doc Al said:
Nope.

You are finding one component of the weight. When finding the component of a vector, the full vector is the hypotenuse. That means mg is the hypotenuse, and the tangential component is one of the sides of the right triangle.

Okay, now I'm really lost. It's been awhile since I've done this stuff.. does this have something to do with mg(parallel), that sort of thing (if I'm making any sense)?
 
  • #28
AJDangles said:
Okay, now I'm really lost. It's been awhile since I've done this stuff.. does this have something to do with mg(parallel), that sort of thing (if I'm making any sense)?
I'm not sure I understand. You are trying to find the forces acting along the tangential direction. The only force with a tangential component is the weight mg. And the angle between the weight vector and the tangent direction is 30°.

How do you find components of a vector at a given angle?
 
  • #29
Doc Al said:
I'm not sure I understand. You are trying to find the forces acting along the tangential direction. The only force with a tangential component is the weight mg. And the angle between the weight vector and the tangent direction is 30°.

How do you find components of a vector at a given angle?

Ok hold on; do I have everything in my diagram labelled correctly? Right now, I am confused as to why you say that mg is the hypotenuse when it's pointing straight down. To me, it looks as if the hypotenuse is the force/component which I have labelled as ma, and that mg would be one of the sides of the right-angle triangle.

Am I looking at the correct triangle here? Or are you talking about a right triangle which is completely different, and one that I haven't shown in my diagram?
 
  • #30
AJDangles said:
Ok hold on; do I have everything in my diagram labelled correctly? Right now, I am confused as to why you say that mg is the hypotenuse when it's pointing straight down. To me, it looks as if the hypotenuse is the force/component which I have labelled as ma, and that mg would be one of the sides of the right-angle triangle.

Am I looking at the correct triangle here? Or are you talking about a right triangle which is completely different, and one that I haven't shown in my diagram?
Yes, the triangle you've drawn is incorrect. (But the direction of the force and tangent line are correct, which is what I was focusing on. Sorry about not pointing that out!) Since you are finding components of the weight, you must draw a new right triangle in which the weight is the hypotenuse.

Sanity check: A vector must always be bigger (or at least equal to) its components. So when drawing a right triangle to find components, the components must be the smaller sides, not the hypotenuse.
 
  • #31
Doc Al said:
Yes, the triangle you've drawn is incorrect. (But the direction of the force and tangent line are correct, which is what I was focusing on. Sorry about not pointing that out!) Since you are finding components of the weight, you must draw a new right triangle in which the weight is the hypotenuse.

Sanity check: A vector must always be bigger (or at least equal to) its components. So when drawing a right triangle to find components, the components must be the smaller sides, not the hypotenuse.

Ok. I drew my triangle with the mg. I found the components to be:

x = mgcos30
y = mgsin30

We're looking for the x component as the acceleration, so then we can go: mgcos30 = ma. Cancel out the masses, solve for acceleration.

a = gcos30
a = 8.5units/s^2

Right? (Please say yes... ha ha)
 
  • #32
AJDangles said:
Ok. I drew my triangle with the mg. I found the components to be:

x = mgcos30
y = mgsin30

We're looking for the x component as the acceleration, so then we can go: mgcos30 = ma. Cancel out the masses, solve for acceleration.

a = gcos30
a = 8.5units/s^2

Right? (Please say yes... ha ha)

I'm thinking this would make sense because at angles of 0 and 180, the acceleration is equal to gravity.
 
  • #33
AJDangles said:
Ok. I drew my triangle with the mg. I found the components to be:

x = mgcos30
y = mgsin30

We're looking for the x component as the acceleration, so then we can go: mgcos30 = ma. Cancel out the masses, solve for acceleration.

a = gcos30
a = 8.5units/s^2

Right? (Please say yes... ha ha)
Right! (or yes, if you prefer!)
 
  • #34
Alright! Last question, I promise: Why is it that the x component gives you the acceleration? I realize that I got the right answer but I'm not so sure as to how I got it. I just assumed it was that way because when you put in cos(0)*9.8 you got 9.8. Which makes sense because the circle is completely vertical at that point.
 
  • #35
AJDangles said:
Alright! Last question, I promise: Why is it that the x component gives you the acceleration?
All you're doing is applying Newton's 2nd law in the tangential direction. For the tangential forces (the x-direction), the only force is the x-component of the weight. So:
ƩFx = max
mg cosθ = max

So... ax = g cosθ

Let me know if this gets at your question. If not, ask again.
 
<h2>1. What is a frictionless half-pipe?</h2><p>A frictionless half-pipe is a curved surface, typically made of metal or plastic, that is completely smooth and has no friction. This means that when an object, such as a skateboard, is placed on the half-pipe, it will not experience any resistance or slowing down due to friction.</p><h2>2. How does a frictionless half-pipe affect rotational motion?</h2><p>A frictionless half-pipe allows for a more ideal or perfect rotational motion. This means that the object will continue to rotate without any external forces acting upon it, such as friction or air resistance. This allows for a more accurate study of rotational motion and its principles.</p><h2>3. What is the significance of studying rotational motion on a frictionless half-pipe?</h2><p>Studying rotational motion on a frictionless half-pipe allows for a better understanding of the fundamental principles of rotational motion. By eliminating the effects of friction, scientists can focus on the effects of other factors, such as mass, velocity, and angular acceleration, on the object's rotation.</p><h2>4. How is a frictionless half-pipe used in scientific experiments?</h2><p>A frictionless half-pipe is often used in experiments to study rotational motion and its principles. Scientists may use it to observe the effects of different factors, such as changing the object's mass or velocity, on its rotation. It can also be used to demonstrate the conservation of angular momentum and other laws of rotational motion.</p><h2>5. Are there any real-life applications of a frictionless half-pipe?</h2><p>While a completely frictionless surface is not possible in the real world, the concept of a frictionless half-pipe can be applied to various real-life situations. For example, in sports such as skateboarding and snowboarding, athletes often perform tricks and maneuvers on half-pipes that are designed to minimize friction and allow for smoother and more consistent rotations.</p>

1. What is a frictionless half-pipe?

A frictionless half-pipe is a curved surface, typically made of metal or plastic, that is completely smooth and has no friction. This means that when an object, such as a skateboard, is placed on the half-pipe, it will not experience any resistance or slowing down due to friction.

2. How does a frictionless half-pipe affect rotational motion?

A frictionless half-pipe allows for a more ideal or perfect rotational motion. This means that the object will continue to rotate without any external forces acting upon it, such as friction or air resistance. This allows for a more accurate study of rotational motion and its principles.

3. What is the significance of studying rotational motion on a frictionless half-pipe?

Studying rotational motion on a frictionless half-pipe allows for a better understanding of the fundamental principles of rotational motion. By eliminating the effects of friction, scientists can focus on the effects of other factors, such as mass, velocity, and angular acceleration, on the object's rotation.

4. How is a frictionless half-pipe used in scientific experiments?

A frictionless half-pipe is often used in experiments to study rotational motion and its principles. Scientists may use it to observe the effects of different factors, such as changing the object's mass or velocity, on its rotation. It can also be used to demonstrate the conservation of angular momentum and other laws of rotational motion.

5. Are there any real-life applications of a frictionless half-pipe?

While a completely frictionless surface is not possible in the real world, the concept of a frictionless half-pipe can be applied to various real-life situations. For example, in sports such as skateboarding and snowboarding, athletes often perform tricks and maneuvers on half-pipes that are designed to minimize friction and allow for smoother and more consistent rotations.

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