Torque acting on a particle in rotational motion

  • #1
peterspencers
72
0

Homework Statement



The following question involves a torque acting on a particle in rotational motion. It provides practice with the various equations for angular velocity, torque etc A particle of mass ##m## initially has position

$$\mathbf{r}=p_{o}[cos\phi(\textit{i}+\textit{k})+sin\phi\sqrt{2}\textit{j}]$$

The angular velocity vector ##\boldsymbol{\omega}##, gives the axis of rotation which is in the direction ##\hat{\mathbf{n}}= \dfrac{− 1}{\sqrt{2}}\textit{i}+\dfrac{1}{\sqrt{2}}\textit{k}##

Gravity acts on the particle, giving rise to a gravitational acceleration -g##\textit{k}## as usual.

Assume that the axis of rotation is fixed in the direction ##\hat{\mathbf{n}}##. Using the component of the torque in the ##\hat{\mathbf{n}}## direction, show that the angular acceleration ##\alpha## about ##\hat{\mathbf{n}}##, as a function of φ is ..

$$\alpha=\dfrac{-gsin\phi}{2p_{o}}$$

Homework Equations

The Attempt at a Solution



$$\textbf{r}=p_{o}\begin{pmatrix}cos\phi\\\sqrt{2}sin\phi\\cos\phi\end{pmatrix}$$
$$r=\sqrt{2}p_{o}$$
$$\textbf{a}=\dfrac{d^{2}\textbf{r}}{d\phi^{2}}+\begin{pmatrix}0\\0\\-g\end{pmatrix}$$
$$\textbf{a}=-p_{o}\begin{pmatrix}cos\phi\\\sqrt{2}sin\phi\\cos\phi-g\end{pmatrix}$$
$$\boldsymbol{\tau}=\textbf{r}\times\textbf{F}=\textbf{r}\times(m\textbf{a})=mr^{2}\boldsymbol{\alpha}$$
so, the angular acceleration about ##\hat{\mathbf{n}}## is $$\alpha=\boldsymbol{\alpha}\cdot\hat{\mathbf{n}}=\dfrac{{\tau}\cdot\hat{\mathbf{n}}}{mr^{2}}$$
(note, I cannot make the greek symbol tau bold (to show that it is a vector) when using the \dfrac command in Latex, how do I achive this ?)

Which I compute to be..

$$\alpha=\boldsymbol{\alpha}\cdot\hat{\mathbf{n}}=\dfrac{-gsin\phi}{2}$$

I seem to be missing a factor of ##\dfrac{1}{p_{o}}## from the solution the question says I should get. Have I made a conceptual error somewhere?
 
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  • #2
In your solution:
1. you interchaged the y and z components of the initial position vector. cording to the problem statement
2. I don't see how you got the acceleration vector. Could you explain that?
 
  • #3
Hi there, I have edited the initial post, apologies, I had put the unit vectors j and k in the wrong place, thankyou for correcting me. I have also included an extra line showing where the acceleration vector comes from, it is simply the second derivative of position with respect to phi + the acceleration due to gravity.
 
  • #4
Why is the acceleration the second derivative of the position with respect to phi? By definition, it is the second derivative of the position with respect to t.
 
  • #5
ok, so the acceleration should just be ##-g\textit{k}##?
 
  • #6
Yes this gives the required result, thankyou very much for taking the time to point out my mistake.

Kind regards,

Peter
 
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