Frictionless half-pipe and rotational motion question

[AJDangles]

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.

 

[Doc Al]

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?)

 

[AJDangles]

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?

 

[Doc Al]

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.)

 

[AJDangles]

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?

 

[Doc Al]

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?

 

[AJDangles]

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?

 

[Doc Al]

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.

 

[AJDangles]

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?

 

[Doc Al]

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.)

 

[AJDangles]

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!

 

[AnanthanK]

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.

 

[AJDangles]

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.

 

[AJDangles]

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

 

[AJDangles]

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)

 

[AnanthanK]

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!

 

[AJDangles]

Thank you very much!

No problem. Do you go to McMaster?

 

[chromium1387]

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!

 

[Doc Al]

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.

 

[Doc Al]

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.

 

[AJDangles]

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.

 

[Doc Al]

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.

 

[AJDangles]

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)

 

[Doc Al]

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.)

 

[AJDangles]

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?

 

[Doc Al]

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.

 

[AJDangles]

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)?

 

[Doc Al]

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?

 

[AJDangles]

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?

 

[Doc Al]

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.

 

[AJDangles]

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)

 

[AJDangles]

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.

 

[Doc Al]

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!)

 

[AJDangles]

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.

 

[Doc Al]

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:
ƩF_x = ma_x
mg cosθ = ma_x

So... a_x = g cosθ

Let me know if this gets at your question. If not, ask again.

 

[AJDangles]

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:
ƩF_x = ma_x
mg cosθ = ma_x

So... a_x = g cosθ

Let me know if this gets at your question. If not, ask again.

Ok... I think I get it now. Thanks a lot. Just one more picture to make sure my triangle is in the right spot: