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The following question is from a past paper from Further Mechanics and it has been bothering me immensely I have spent hours cracking at a way to make sense of it, the question is in two parts

Part 1:

An object consists of a uniform solid circular cone, of vertical height 4r and radius 3r, and a uniform solid cylinder, of height 4r and radius 3r. The circular base of the cone and one of the circular faces of the cylinder are joined together so that they coincide. The cone and the cylinder are made of the same material.

Find the distance of the centre of mass of the object from the end of the cylinder that is not attached to the cone.

I first roughly sketched the object as such

sketch

Then finding the $\bar{x}$ of each part seperately:

For the cone I used the standard result of the centre of mass being $\frac{1}{4}r$ away from the base:

$$\bar{x}_{Cone}=\frac{1}{4}.4r + 4r=5r$$

For the cylinder I derived $\bar{x}$ via integration:

$$y=3r$$

$$\bar{x}_{cylinder}=\frac{\int_0^{4r} xdV}{\int_0^{4r}dV}$$

$$\because dV=y^2 \pi dx $$

$$\implies \bar{x}_{cylinder}=\frac{\int_0^{4r} 9r^2 \pi x dx}{\int_0^{4r}9r^2 \pi dx}$$

$$\implies \bar{x}_{cylinder}=\frac{9r^2 \pi \int_0^{4r} x dx}{9r^2 \pi\int_0^{4r} 1 dx}$$

$$\implies \bar{x}_{cylinder}=\frac{\frac{1}{2}\left[x^2 \right]^{4r}_0}{\left[x \right]^{4r}_0}$$

$$\therefore \bar{x}_{cylinder}=2r $$

Then by taking the weighted average of both objects I arrived at the correct value for the distance of the centre of gravity from the base of the cylinder which was

$$\bar{x}=\frac{11}{4}r$$

However the next part has had me at a complete loss for the better part of the day, it states that:

Show that the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface.

I tried coming up with a rough sketch for this also

sketch 2

However I do not understand how to tackle this question at all, all I know is that for a body to be at equilibrium on such a surface, the centre of mass must pass through the point of suspension, however clearly this does not give an insight on how to answer this.

The condition for this question given in the marking scheme is that

markscheme

Can someone explain what this means and why this is the case?

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Given the geometry of the object, the center of gravity is very close to the surface between the cylinder and the cone. More specifically the distance between the center of gravity and the base of the cone is $\frac{1}{4}r$.

Basically what you need to prove is that the angle of the cone $ \phi$ is such, so that when the object is tilted the weight crosses over the last $\frac{1}{4}r$.

enter image description here

So you need to prove that the angle $\theta$ is smaller that angle $\phi$.

Angle $\theta$ is calculated as: $$\tan\theta = \frac{\text{distance to be covered}}{\text{cylinder radius}}= \frac{\frac{11}{4}r}{3r}=\frac{11}{12} \Rightarrow \theta =42.5 [deg] $$

while Angle $\phi$ is calculated as: $$\tan\phi= \frac{\text{height of cone}}{\text{cone radius}}= \frac{4r}{3r}=\frac{4}{3} \Rightarrow \phi =53.1[deg] $$

Therefore, since $\phi> \theta$, the object becomes vertical enough so that the weight passes through the coned surface.

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    $\begingroup$ @Filthyscrub Apologies for my initial answer, I misread your question. $\endgroup$
    – NMech
    Mar 23 at 9:52
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    $\begingroup$ Reminds me of a crafty problem my undergrad QM prof posed. I'll adapt it to this particular object: using Heisenberg's Uncertainty Principle, how long will this object remain stable resting on the cone? It's not intended as a "real-world" problem but rather to see how you might use paired-parameters to guesstimate the stability. $\endgroup$ Mar 24 at 12:43
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The CG is found by assuming the mass of the cone as 1/3m and the cylinder m, then

$$ \bar{X}= \frac{2r*m+5r*m/3}{4/3m} =r*11/4 $$

The cone side length (like the sharp tip of a pencil) is $5r$, the side of a 3,4,5 triangle, and the interior half tip angle is 36.87 degrees.

The CG is $5.25r$ from the tip of the cone and at rotation to rest on the side of the cone will be at $$5.25r*cos36.87= 4.19r<5r$$

I let you prove the condition.

It is well within the footprint of the pencil shape so it is stable.

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