# Angular velocity equation

## Homework Statement

A uniform cylindrical wheel of mass $m_{1}$ and radius $R_{1}$ rotates with angular velocity $\omega_{1}$. It lies a certain distance (along the same axis) from a static wheel of radius $R_{2}$ and mass $m_{2}$. The wheels are then pushed against each other with a constant force $F$ uniformly distributed across the wheel's face. There is friction between the wheels.
What is the final angular velocity of the two wheels $\omega_{f}$ and how long does it take for the wheels to reach that speed? ## Homework Equations

$\tau=I\ddot{\omega}$, maybe conservation laws?

## The Attempt at a Solution

There is no conservation of angular momentum since the torque is changing.
The initial torque is $\tau=\frac{2}{3}R_{1}F\mu$. I have to find $I$ for the cylinders. How do I relate $\tau$ at a given time with $\omega$?

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haruspex
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There is no conservation of angular momentum since the torque is changing.
Angular momentum about a given axis is conserved for a system if there are no external forces that exert a torque about that axis.
What are you defining as the system? What external forces act on that system? Do they exert a torque about your chosen axis?

However, since you have to find the time taken, using a conservation law won't solve it. You have to consider the torque each disc exerts on the other.
Consider an annular element radius r, width dr, of the contact surfaces. What share of the normal force is exerted over that region. What torque does that imply?

• Jenny Physics
Angular momentum about a given axis is conserved for a system if there are no external forces that exert a torque about that axis.
What are you defining as the system? What external forces act on that system? Do they exert a torque about your chosen axis?

However, since you have to find the time taken, using a conservation law won't solve it. You have to consider the torque each disc exerts on the other.
Consider an annular element radius r, width dr, of the contact surfaces. What share of the normal force is exerted over that region. What torque does that imply?
The torque exerted by disk 1 on disk 2 is $d\tau_{1}=r\frac{F}{\pi R_{1}^{2}}2\pi rdr \tau_{1}=\int_{0}^{R_{1}}=\frac{2}{3}R_{1}F$ the torque exerted by disk 2 on disk 1 will then be $\tau_{2}=\frac{2}{3}R_{2}F$. These are constant torques and that would lead to an angular momentum that varies linearly with $t$?

haruspex
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The torque exerted by disk 1 on disk 2 is $d\tau_{1}=r\frac{F}{\pi R_{1}^{2}}2\pi rdr$, $\tau_{1}=\int_{0}^{R_{1}}=\frac{2}{3}R_{1}F$ the torque exerted by disk 2 on disk 1 will then be $\tau_{2}=\frac{2}{3}R_{2}F$. These are constant torques and that would lead to an angular momentum that varies linearly with $t$?
You left out the friction coefficient, and you need to think more about the significance of the radii maybe being different. can the torques they exert on each other be different?
Yes, the angular acceleration will be constant.

You left out the friction coefficient, and you need to think more about the significance of the radii maybe being different. can the torques they exert on each other be different?
Yes, the angular acceleration will be constant.
If the angular acceleration is constant that means the angular velocity will decrease with time and there is no $\omega_{f}$?
If I look at the torque on disk 2 due to disk 1 it is $\tau_{21}=\frac{2}{3}F(R_{1}-R_{2}+\mu)$

haruspex
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If the angular acceleration is constant that means the angular velocity will decrease with time and there is no $\omega_{f}$?
What happens to a kinetic friction force when the two surfaces reach the same velocity?
If I look at the torque on disk 2 due to disk 1 it is $\tau_{21}=\frac{2}{3}F(R_{1}-R_{2}+\mu)$
How do you get that? It is not dimensionally consistent.

• Jenny Physics
What happens to a kinetic friction force when the two surfaces reach the same velocity?

How do you get that? It is not dimensionally consistent.
The friction force becomes zero. But how do I model a changing friction force?
Typo. I meant $\tau_{21}=\frac{2}{3}F(R_{1}-R_{2}+\mu R_{1})$

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haruspex
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The friction force becomes zero. But how do I model a changing friction force?
Same way as for linear motion. The SUVAT equations map across to rotational motion quite simply.
Typo. I meant $\tau_{21}=\frac{2}{3}F(R_{1}-R_{2}+\mu R_{1})$
Still wrong. If the coefficient is zero the torque will be zero.
If the radii are different, where does the friction occur?

• Jenny Physics
Same way as for linear motion. The SUVAT equations map across to rotational motion quite simply.

Still wrong. If the coefficient is zero the torque will be zero.
If the radii are different, where does the friction occur?
Not following. How do I model the jump from $\mu$ constant to a zero friction coefficient?
If the radii are different friction occurs on the circle of the smaller radius disk?

haruspex
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If the radii are different friction occurs on the circle of the smaller radius disk?
Yes. So what equation do you get?
How do I model the jump from $\mu$ constant to a zero friction coefficient?
How do you find the time for a braking car to come to a stop?
But in this case, of course, it is relative motion that ceases.

• Jenny Physics
Yes. So what equation do you get?
$\frac{2}{3}\mu FR_{1}$ assuming $R_{1}$ is the smaller radius.

How do you find the time for a braking car to come to a stop?
But in this case, of course, it is relative motion that ceases.
In this case I have to find the time at which the two wheels have the same angular velocity $\omega_{f}$, so I assume I have to write two equations one for cylinder 1 and the other for cylinder 2 using $\tau_{1}=I_{1}\dot{\omega}_{1},\tau_{2}=I_{2}\dot{\omega}_{2}$

haruspex
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$\frac{2}{3}\mu FR_{1}$ assuming $R_{1}$ is the smaller radius.

In this case I have to find the time at which the two wheels have the same angular velocity $\omega_{f}$, so I assume I have to write two equations one for cylinder 1 and the other for cylinder 2 using $\tau_{1}=I_{1}\dot{\omega}_{1},\tau_{2}=I_{2}\dot{\omega}_{2}$
Yes, the relationship between the two torques being...?

• Jenny Physics
Yes, the relationship between the two torques being...?
The two torques will be equal but opposite in sign i.e. $\tau_{1}=-\tau_{2}$. So the equations of motion will be $-\frac{2}{3}\mu FR_{1}=\frac{1}{2}mR_{1}^{2}\dot{\omega_{1}},\frac{2}{3}\mu FR_{1}=\frac{1}{2}mR_{2}^{2}\dot{\omega_{2}}$ (notice how the moment of inertia are different because the radii are different)

haruspex
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The two torques will be equal but opposite in sign i.e. $\tau_{1}=-\tau_{2}$. So the equations of motion will be $-\frac{2}{3}\mu FR_{1}=\frac{1}{2}mR_{1}^{2}\dot{\omega_{1}},\frac{2}{3}\mu FR_{1}=\frac{1}{2}mR_{2}^{2}\dot{\omega_{2}}$ (notice how the moment of inertia are different because the radii are different)
Almost.
What is the final angular velocity of wheel 1?

• Jenny Physics
Where ω1 is...?
$\omega_{1}$ will be found by integration so should be $\omega_{1}(t)=\omega_{1}(0)-\frac{4}{3mR_{1}}\mu F t$ and $\omega_{2}(t)=\frac{4}{3mR_{2}}\mu Ft$. Setting them equal gives the time $t$

haruspex
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$\omega_{1}$ will be found by integration so should be $\omega_{1}(t)=\omega_{1}(0)-\frac{4}{3mR_{1}}\mu F t$ and $\omega_{2}(t)=\frac{4}{3mR_{2}}\mu Ft$. Setting them equal gives the time $t$
Not quite.
You made the assumption that R1 is smaller and in post #11 wrote the torque that applies to both. You need to avoid that assumption. I suggest you define R = min{R1, R2} and write the torque in terms of that.

Note that whichever is smaller, one of the expressions in post #15 is wrong.

• Jenny Physics
Not quite.
You made the assumption that R1 is smaller and in post #11 wrote the torque that applies to both. You need to avoid that assumption. I suggest you define R = min{R1, R2} and write the torque in terms of that.

Note that whichever is smaller, one of the expressions in post #15 is wrong.
Ok assuming $R_{1}$ to be the smaller, $\omega_{1}$ will be found by integration so should be $\omega_{1}(t)=\omega_{1}(0)-\frac{4}{3mR_{1}}\mu F t$ and $\omega_{2}(t)=\frac{4R_{1}}{3mR_{2}^{2}}\mu Ft$. Setting them equal gives the time $t$

haruspex
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Ok assuming $R_{1}$ to be the smaller, $\omega_{1}$ will be found by integration so should be $\omega_{1}(t)=\omega_{1}(0)-\frac{4}{3mR_{1}}\mu F t$ and $\omega_{2}(t)=\frac{4R_{1}}{3mR_{2}^{2}}\mu Ft$. Setting them equal gives the time $t$
Yes, but you cannot arbitrarily take R1 to be smaller because R1 is already defined as the one initially rotating at the given rate.
You can either provide two answers, one for each possibility, or use the expression $\min\{R_1, R_2\}$ in the answer.

• Jenny Physics
Yes, but you cannot arbitrarily take R1 to be smaller because R1 is already defined as the one initially rotating at the given rate.
You can either provide two answers, one for each possibility, or use the expression $\min\{R_1, R_2\}$ in the answer.
I agree. I have one lingering question which is why can we cancel the torques due to F on a given wheel? I agree that when $R_{1}=R_{2}$ the torques are equal in magnitude and opposite, but since the torque is an integral of $d\tau=r\frac{F}{\pi R^{2}}2\pi r dr$ will this not imply that the torque of the larger cylinder is larger than the torque of the smaller cylinder?

haruspex
I agree. I have one lingering question which is why can we cancel the torques due to F on a given wheel? I agree that when $R_{1}=R_{2}$ the torques are equal in magnitude and opposite, but since the torque is an integral of $d\tau=r\frac{F}{\pi R^{2}}2\pi r dr$ will this not imply that the torque of the larger cylinder is larger than the torque of the smaller cylinder?
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