An Eskimo child sliding down an igloo

[ChessEnthusiast]

Homework Statement

A child of an Eskimo decided to slide down an igloo. The igloo looks like a hemisphere with radius R. When will the child fall off the slide?

Homework Equations

See below

The Attempt at a Solution

There are only two forces acting on the child - the force of gravity and the force of friction. I assume that in any given point the child is in fact sliding down a line - tangent to the circle. Therefore, the forces are:
$$F_n = mg \cos(\theta) - \mbox{normal to the igloo} \\ F_t = mg \sin(\theta) - \mbox{tangent to the circle} \\ F_f = \mu mg \cos(\theta) - \mbox{friction}$$
$$\overrightarrow{F_n} + \overrightarrow{F_t} = m \overrightarrow{g}$$

Now, I am not sure about this but the normal force (the component of the force of gravity) seems to function as the centripetal force, therefore
this is how I would proceed with solving this problem: <br>
I would use the tangent force to express velocity of the child as function of time. Then, I would compare the normal force and the centripetal force to find the time when the required centripetal force will be larger than the normal force - then, the child will fall off.

Is this a correct way to solve this?

 

[BvU]

There are only two forces acting on the child

Yes

the force of gravity and the force of friction

No. Assume there is no friction

 

[ChessEnthusiast]

No. Assume there is no friction

Why so?
The second (relatively third) is the force of reaction due to the 3rd law of motion.

 

[BvU]

The second is the normal force the iglo exerts on the child - it does not accelerate "inward", but follows the surface if the igloo, until...

 

[ChessEnthusiast]

Does the fact that the child follows the circular surface of the igloo mean that there is no centripetal acceleration?

 

[BvU]

The child follows a circular trajectory, so there must be a centripetal force. The provider ?

 

[ChessEnthusiast]

The radial component of the force of gravity?

 

[BvU]

Correct

 

[ChessEnthusiast]

As you said, I have decided to ignore friction. Therefore I can express acceleration and velocity as function of the angle:
$$a(\theta) = g \sin (\theta)$$
$$v(\theta) = -g \cos(\theta)$$
The radial component of the force of gravity is the centripetal force:
$$\frac{m(v(\theta))^2}{R} = mg \cos(\theta)$$
Thus
$$cos(\theta) = \frac{R}{g}$$
Correct?
Also, I can solve for theta with respect to x and R, since
$$x^2 + y^2 = R^2 \Longrightarrow \frac{dy}{dx} = - \frac{x}{y}$$

 

[BvU]

The radial component of the force of gravity is the centripetal force

Not when e.g. $\theta = 0$ !

 

[ChessEnthusiast]

Do we really need to care about this initial stage when $$\theta = 0$$? If so, where did my reasoning go wrong?

 

[BvU]

Do we really need to care about this initial stage when

θ=0​

Not really. It is a so-called unstable equilibrium position.
But: It is important to realize the radial gravity component is initially greater than the force that is required to follow a circular trajectory. It is only equal to that at the point where the kid leaves the surface of the igloo.

There are a few problems with post #9: $v(\theta) = -g \cos(\theta)$ ?? Wrong dimension. Then $cos(\theta) = \frac{R}{g}$: same thing...not to mention that the left side should not be independent of time !

 

[ChessEnthusiast]

I have spent some more time thinking about my approach and I think that I am going to have a hard time trying to solve it this way.
Namely, the function of angle with respect to time depends on the angle...:
$$\theta(t) = \frac{g}{R} \sin(\theta(t))$$
I have decided to try something different, namely, we know that the speed our object has to reach is defined by the radial component of the force of gravity, namely
$$mg\cos(\theta) = m \frac{v^2}{R} \Rightarrow v^2 = Rg \cos(\theta)$$
Therefore, we need to reach this speed. <br>
I assume to friction, therefore <br>
$$mgR = \frac{1}{2}m(Rg \cos(\theta)) + mgh$$
where h is the height on which the child will leave the surface of the igloo. <br>
What about this approach?

 

[PeroK]

I have spent some more time thinking about my approach and I think that I am going to have a hard time trying to solve it this way.
Namely, the function of angle with respect to time depends on the angle...:
$$\theta(t) = \frac{g}{R} \sin(\theta(t))$$
I have decided to try something different, namely, we know that the speed our object has to reach is defined by the radial component of the force of gravity, namely
$$mg\cos(\theta) = m \frac{v^2}{R} \Rightarrow v^2 = Rg \cos(\theta)$$
Therefore, we need to reach this speed. <br>
I assume to friction, therefore <br>
$$mgR = \frac{1}{2}m(Rg \cos(\theta)) + mgh$$
where h is the height on which the child will leave the surface of the igloo. <br>
What about this approach?

So far, so good. Although you've forgotten to square the speed in your energy equation.

 

[ChessEnthusiast]

I haven't, $$[Rg] = \frac{meters^2}{seconds^2}$$

 

[ChessEnthusiast]

So, let's say that I want to continue solving this. I have arrived at this equation:
$$mgR = \frac{1}{2}m(Rg \cos(\theta)) + mgh$$
$$R = \frac{1}{2}R \cos(\theta) + h$$

This equation contains two variables. Is there a better way of disposing of one of them than using this identity:
$$\frac{dy}{dx} = - \frac{x}{y}$$?

 

[PeroK]

So, let's say that I want to continue solving this. I have arrived at this equation:
$$mgR = \frac{1}{2}m(Rg \cos(\theta)) + mgh$$
$$R = \frac{1}{2}R \cos(\theta) + h$$

This equation contains two variables. Is there a better way of disposing of one of them than using this identity:
$$\frac{dy}{dx} = - \frac{x}{y}$$?

What answer are you looking for, $h$ or $\theta$? In either case, don't you have enough equations?

Note that the time is indeterminate for this problem. You might like to think why this is the case.

 

[kuruman]

This equation contains two variables. Is there a better way of disposing of one of them than using this identity:

Yes. Make a drawing and express $h$ in terms of $R$ and $\theta$. Then the radius drops out. In short, the angle of separation is independent of the size of the igloo.