1.A solid cone of length \(h\) and half angle \(\alpha\) is rolling on a plane about its vortex with an angular velocity of \(\Omega\hat{k}\), as shown in Fig. Compute the angular velocity, angular momentum, and kinetic energy of the cone.

2. The vertex of the aforementioned cone is fixed on the \(z\) axis at a height equal to the radius of the cone. The cone rotates an angular velocity of \(\Omega\hat{k}\) about the vertical axis as shown in Fig. Compute the angular velocity, angular momentum, and kinetic energy of the cone.
No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestHere is my attempt at the second one. There are two rotations to consider:

1)The rotation of the cone about its own axis2)The rotation of the cone about the vertical axisLet \(H\) be the height of the cone, \(R\) be the radius of the cone, and \(\alpha\) be the semi-angle.

\[tan \, \alpha = \frac{R}{H} \\ H = \frac{R}{tan \, \alpha}\]

Let the angular speed of the cone with respect to the vertical axis be \(\omega\), and the angular speed of the cone with respect to its own axis be \(\omega'\). They are related as follows (assuming no slipping):

\[H \, \omega = R \, \omega'\]

Let \(I\) and \(I'\) be the moments of inertia with respect to the vertical axis and the cone axis, respectively.

\[I = \frac{3}{20} \, M \, (R^2 + 4 H^2) \\ I' = \frac{3}{10} \, M \, R^2 \]

The kinetic energy is then (angular momentum is similar):

\[E = \frac{1}{2} I \, \omega^2 + \frac{1}{2} I' \, \omega'^2 \\ \frac{3}{40} \, M \, (R^2 + 4 H^2) \, \omega^2 + \frac{3}{20} \, M \, R^2 \, \omega'^2 \\ = \frac{3}{40} \, M \, R^2 \, \Big( 1 + \frac{4}{tan^2 \, \alpha} \Big) \, \omega^2 + \frac{3}{20} \, M \, R^2 \, \frac{1}{tan^2 \, \alpha} \, \omega^2 \\ = \frac{3}{40} \, M \, R^2 \, \omega^2 + \frac{9}{20} \, M \, R^2 \, \frac{1}{tan^2 \, \alpha} \, \omega^2 \]

Log in to reply

Thx @Steven Chase sir got.it.(How to.solve first one any ideas ? As such axis of rotation would.be at an angle.from angular.momentum vector right )??

Log in to reply

I'll start thinking about the first one. Did I get the second one right?

Log in to reply

@Steven Chase Sir ( David morin has the first part of the problem in it , but I dont know how they approach the question) . ( they confused me by adding angular momentums )

Yes u r rightLog in to reply

Log in to reply

@Steven Chase sir pls share how ur way of solving these 2 problem ? @Aaghaz Mahajan bro pls help u also ?

Log in to reply

Do you know the answers? I think the second one should be fairly easy. The first one will take a bit more work.

Log in to reply

@Steven Chase sir, Can u solve for second part and later first part sir? (Yes I know answers for second part .)

Log in to reply

Log in to reply