Toboggan on circular arcs

[Physical_Fire]

Hi,
I have a homework question involving conservation of energy and centripetal force. My instructor provided an example with a solution, which I tried to apply to my homework problem, but without success. I tried using V as the initial speed, but in the end I ended up with a range for V. Could someone guide me in the right direction?

Thanks

 

[kuruman]

We cannot guide you in the right direction without knowing how and where you got lost. Please post you attempt at a solution and the specific problem you were asked to solve.

 

[Physical_Fire]

Here’s what I did and got:

 

[kuruman]

Please show some consideration to those who are trying to help you. We are not mind readers.

What is the statement of the problem?
Also a diagram would be helpful. What are points A, B and C?

 

[Physical_Fire]

What is the statement of the problem?

What do you mean? The question is attached above.

Also a diagram would be helpful. What are points A, B and C?

They are symmetrical as far as I can see.

 

[kuruman]

What do you mean? The question is attached above.

Yes, sorry I confused your problem with the example. In your solution, why do you say that the potential energy at A is mg(20)? Where do you take the potential energy to be zero?

 

[Physical_Fire]

Where do you take the potential energy to be zero?

At the very top; the centre of the circle (D).

 

[kuruman]

At the very top; the centre of the circle (D).

I think that would make thing difficult on the other side of the equation. To begin with, the potential energy at the top of the arc between B and C should negative because the top of the arc is below point D. Secondly, the vertical distance between D and the top of the arc might be a bit tricky to calculate.

I suggest that you choose the potential energy to be zero at the horizontal line from A to C. Then all you have to find is the vertical distance from that line to the top of the arc.

On edit
In the example the zero of potential energy is the centre of the arc on the right at point E in your drawing. You could choose that but then you have to have the vertical distance from E to line AC in order to find the initial potential energy.

Also, I think that the solution provided in the example is incorrect. The term $\cos\theta$ does not belong in the $F=ma$ equation provided in the example (see screenshot on the right). It is written when the mass is at the top of the trajectory, where the net force is the weight $mg$ (without the cosine) and the normal force $R$. The positive direction is assumed "down."

 

[Physical_Fire]

I suggest that you choose the potential energy to be zero at the horizontal line from A to C. Then all you have to find is the vertical distance from that line to the top of the arc.

Alright. I’ll try doing that but as before I get my answer in terms of V. How do I get in terms of U?
And I think you misread the problem. In the example problem, U is the initial velocity (so mgcos(θ) makes sense) and not the midpoint like in the problem I have to solve.

 

[haruspex]

energy.

Also, I think that the solution provided in the example is incorrect. The term cos⁡θ does not belong in the F=ma equation provided in the example (see screenshot on the right). It is written when the mass is at the top of the trajectory, where the net force is the weight mg (without the cosine) and the normal force R. The positive direction is assumed "down."

Correctly, the analysis considers the highest risk of losing contact is at A and B. F=ma is being applied along the normal, so at 45° to the vertical.

 

[kuruman]

Correctly, the analysis considers the highest risk of losing contact is at A and B. F=ma is being applied along the normal, so at 45° to the vertical.

Yes, I ignored that it's the point where the centripetal acceleration changes direction.

 

[haruspex]

I tried using V as the initial speed, but in the end I ended up with a range for V. Could someone guide me in the right direction?

I see no benefit in introducing V. You are to find a range of values for U
First, identify the points of the trajectory which will determine the minimum and maximum speeds and what the constraints are on the speeds at those points. Then use energy conservation to relate those to U.

At the very top; the centre of the circle (D).

As @kuruman notes, that means the PE at each point will be negative, but you have written them as positive.

 

[kuruman]

Alright. I’ll try doing that but as before I get my answer in terms of V. How do I get in terms of U?

What is in a symbol? If U is the speed at the lowest point of the motion, you are looking for its maximum and minimum values, call them $U_{\text{max}}$ and $U_{\text{min}}$ such that $U_{\text{min}}<U<U_{\text{max}}.$