Torque acting on a particle in rotational motion

AI Thread Summary
The discussion revolves around a problem involving torque acting on a particle in rotational motion, specifically focusing on deriving the angular acceleration as a function of the angle φ. The particle's position and the direction of the angular velocity vector are defined, with gravity affecting the motion. A calculation error was identified regarding the missing factor of 1/p_o in the angular acceleration formula. The conversation clarified the correct derivation of the acceleration vector and addressed a mix-up in the component placement of the initial position vector. Ultimately, the participants resolved the conceptual misunderstandings and confirmed the correct approach to the problem.
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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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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?
 
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.
 
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.
 
ok, so the acceleration should just be ##-g\textit{k}##?
 
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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