Ratio of rotational inertias for belt-connected wheels

[kuuy]

Homework Statement

Wheels A and B as shown in the figure are connected by a massless belt that does not slip. The radius of A is R and the radius of B is r. What is the ratio of rotational inertias $\frac{I_a}{I_b}$ if the two wheels had the same angular momentum about their central axes?

Figure:

Multiple choice:

Homework Equations

$I = \frac{rf}{a}$

The Attempt at a Solution

The rotational inertia of A would be $I_a = \frac{Rf}{a} = Rf\frac{1}{a}$
The rotational inertia of B would be $I_b = \frac{rf}{a} = rf\frac{1}{a}$

The $f$ and the $\frac{1}{a}$ both cancel out.
Thus, $\frac{I_a}{I_b}$ would be $\frac{R}{r}$

This is correct according to the answer key (the answer was A). Is this the right way to go about it, or am I getting my answer by sheer luck? Where does the equal angular momentum come into play?

Thanks!

 

[kuruman]

Where did you get your expression for the rotational inertia? What do the symbols f and a mean? You should use the given information that the two wheels have the same angular momentum. Apart from that, what else do they have that is the same?

 

[gneill]

Hi kuuy,

Welcome to Physics Forums.

Please note that I've changed your thread title to better describe the actual problem. Titles must be highly descriptive of the problem so that helpers can quickly identify the ones that pertain to their areas of expertise.

 

[kuuy]

Hi kuuy,

Welcome to Physics Forums.

Please note that I've changed your thread title to better describe the actual problem. Titles must be highly descriptive of the problem so that helpers can quickly identify the ones that pertain to their areas of expertise.

Thank you! That was close to my original title, but then I changed it because of the nature of my question!

 

[gneill]

The rotational inertia of A would be $I_a = \frac{Rf}{a} = Rf\frac{1}{a}$
The rotational inertia of B would be $I_b = \frac{rf}{a} = rf\frac{1}{a}$

The $f$ and the $\frac{1}{a}$ both cancel out.
Thus, $\frac{I_a}{I_b}$ would be $\frac{R}{r}$

You would have to demonstrate that both wheels have the same angular acceleration, α, for your proof to be valid. Given the scenario in the drawing, can you make that claim?

 

[kuuy]

Where did you get your expression for the rotational inertia? What do the symbols f and a mean? You should use the given information that the two wheels have the same angular momentum. Apart from that, what else do they have that is the same?

I am using Newton's Second Law for rotational motion, $rf = I\alpha$, where $r$ is the radius, $f$ is the frictional force, and $\alpha$ is the angular acceleration. By writing the equation in terms of I, I am able to get rid of $f$, and $\alpha$, leaving me with just $\frac{R}{r}$.

 

[gneill]

I am using Newton's Second Law for rotational motion, $rf = I\alpha$, where $r$ is the radius, $f$ is the frictional force, and $\alpha$ is the angular acceleration. By writing the equation in terms of I, I am able to get rid of $f$, and $\alpha$, leaving me with just $\frac{R}{r}$.

I understand what you are trying to do, but you haven't shown that the same α applies to both wheels. You can argue that the force is the same, since it's due to any tension in the belt, but you need to show that the acceleration is the same before you are allowed to cancel the values.

Start by writing your Newton's law equation for each wheel. Use different variables for each:

$R f_a = I_a α_a$
$r f_b = I_b α_b$

Which values are the same for each equation?

 

[kuruman]

If the problem is to be addressed using torques under the assumption that there is angular acceleration, then the changing angular momentum must come from an external source, e.g. a motor, driving one of the wheels. The two-wheel system cannot accelerate all by itself. In that case one should write, assuming that A is the driven wheel,
$\tau_{motor}-Rf_{on~A~by~B} = I_A \alpha_{A}$
$rf_{on~B~by~A} = I_B \alpha_{B}$
Without knowing the torque supplied by the motor or one of the accelerations, Newton's 2nd law doesn't lead anywhere. If there is no angular acceleration, then Newton's 2nd law doesn't lead anywhere either.

I suggest that you start with the angular momentum equations
$L_A=I_A \omega_A$
$L_B=I_B \omega_B$
and see what happens when the two angular momenta are equal as the problem requires.