Rotational Mechanics Question - A Rotating Bar

[Mayhem]

Homework Statement

A bar of mass m is attached to a pivot point and creates an angle φ with horizontal. It moves counter-clockwise from the initial angle, φ_0, to the final angle, φ_1. What is the total work done by the bar?

Relevant Equations

F=mg*cos(φ)

This isn't really a proper homework question, but something I wondered about myself. To simplify things, we say that pivot point is in the origin of a Cartesian coordinate system, and the angles are constrained to the first quadrant.

We see that the weight of the barbell is given as $$F = mgcos(\phi)$$. Since in translatoral mechanics, we define work as $$W = \int_{x_0}^{x_1} ma dx$$, would it make sense in rotational mechanics to define work as $$W = \int_{\phi_0}^{\phi_1} ma\cos{\phi} d\phi$$ ?

In that case, I get the general solution of the work being done as

$$W = mg(\sin{\phi_1}-\sin{\phi_0})$$

 

[etotheipi]

What is the total work done by the bar?

In this case there are no other objects in the system for the bar to do work on, so it would be better to speak of the work done on the bar.

We see that the weight of the barbell is given as $$F = mgcos(\phi)$$

This is the component of weight perpendicular to the bar.

would it make sense in rotational mechanics to define work as $$W = \int_{\phi_0}^{\phi_1} ma\cos{\phi} d\phi$$

Not quite. Firstly, instead of $ma$, try to use $F$ here. The $a$ you have used in this context is not well defined.

The work always goes like $dW = \vec{F} \cdot d \vec{r}$, i.e. the component of the force in the $d \vec{r}$ direction times the magnitude of the displacement. Consider the motion of the centre of mass, through which the weight force acts, if the rod rotates by $d \phi$. What is the size of the arc length traced out? What is the work done?

 

[Mayhem]

In this case there are no other objects in the system for the bar to do work on, so it would be better to speak of the work done on the bar.
This is the component of weight perpendicular to the bar.

Not quite. Firstly, instead of $ma$, try to use $F$ here. The $a$ you have used in this context is not well defined.

The work always goes like $dW = \vec{F} \cdot d \vec{r}$, i.e. the component of the force in the $d \vec{r}$ direction times the magnitude of the displacement. Consider the motion of the centre of mass, through which the weight force acts, if the rod rotates by $d \phi$. What is the size of the arc length traced out? What is the work done?

Calculating the arc length would just give a distance. Could I then calculate the work done on the bar with #W = Fr#?

I originally thought about this, but didn't know if it would work.

 

[etotheipi]

Yes, dimensional analysis can tell you when a formula is wrong without even seeing the derivation.

Calculating the arc length would just give a distance. Could I then calculate the work done on the bar with #W = Fr#?

That's right, though you can still use your integral construction with few changes. If a force F acts in a perpendicular direction to the rod, then the work done is $dW = Fr d\phi$. Have a look at the term $Fr$; does it remind you of any other quantity? Specifically, what do you get if you multiply a force by a perpendicular distance to a point?

 

[kuruman]

Homework Statement:: A bar of mass m is attached to a pivot point and creates an angle φ with horizontal. It moves counter-clockwise from the initial angle, φ_0, to the final angle, φ_1. What is the total work done by the bar?

This problem is ill-posed. Without mention of a force or torque or linear or angular acceleration, the work done on or by the bar is undefined.

 

[Mayhem]

Yes, dimensional analysis can tell you when a formula is wrong without even seeing the derivation.That's right, though you can still use your integral construction with few changes. If a force F acts in a perpendicular direction to the rod, then the work done is $dW = Fr d\phi$. Have a look at the term $Fr$; does it remind you of any other quantity? Specifically, what do you get if you multiply a force by a perpendicular distance to a point?

You get the torque. Which due to obvious reasons reflects the work done on the bar. I see it now. So I can use $dW = Fr d\phi$ for constant circular motion?

 

[haruspex]

You get the torque. Which due to obvious reasons reflects the work done on the bar. I see it now. So I can use $dW = Fr d\phi$ for constant circular motion?

It depends what you know.

If you know that a force is applied at a particular point causing a rotation through some angle (these need not be constant, let them be functions of the angle) then for the work the force does you can write $dW=(\vec r(\phi)\times\vec F(\phi)).\vec{d\phi}$, where $\vec r$ is the vector from the axis to the point of application.
Note that the elemental change in angle is a vector. We can also make everything a function of time, if appropriate: $dW=(\vec r(t)\times\vec F(t)).\vec{\omega}dt$

But going back to your original phrasing, you might complete the description by saying that the bar has some initial rotation rate and is only subject to gravity and the support force from the axle. In this case you could ask what work the bar does (i.e., against gravity) and express it in terms of the mass and the change in elevation of the mass centre.

 

[archaic]

I should have drawn $\vec g$ at the center of the bar, but since it is vector and we're taking its scalar product with $d\vec l$, nothing will go wrong.
$$W=\int_{y_0}^{y_1}-mg\underbrace{\cos\phi\,dl}_{dy}=\int_{y_0}^{y_1}-mg\,dy=-mg\Delta y$$
If you let $y(\phi)=L\sin\phi$, then $dy=L\cos\phi\,d\phi$, and the bounds are $\phi_0=\arcsin\frac{y_0}{L}$, and $\phi_1=\arcsin\frac{y_1}{L}$.
Or you can just infer that, since $l=L\phi$, $dl=Ld\phi$.

 

[wrobel]

Let $A$ be an arbitrary point of a rigid body. Assume that the body experiences forces with the net force $\boldsymbol F$ and the net torque $\boldsymbol M_A$ and let $W$ be the work done by these forces
then$$\dot W=(\boldsymbol F,\boldsymbol v_A)+(\boldsymbol \omega,\boldsymbol M_A)$$

 

[etotheipi]

This also applies to general rigid body rotation, and does not require the rod to be rotating about a fixed axis. Choosing any arbitrary coordinate system you can say $$W = \int \vec{\tau} \cdot d\vec{\theta}$$ so long as you know the torque as a function of the angular displacement of the point of application of the force (you can prove it by letting $d\vec{r} = d\vec{\theta} \times \vec{r}$, since it's essentially just a restatement of the old expression for work).