Oscillatory motion problem: Bicycle wheel rotation oscillation

[Brian_D]

Homework Statement

A bicycle is turned upside down and the front wheel executes a slow, small-amplitude back-and-forth rotational motion with a period of 12 s. Considering the wheel to be a thin ring of mass 600 g and radius 30 cm, whose only irregularity is the presence of the tire valve stem, determine the mass of the valve stem.

Relevant Equations

rotational inertia (I) for simple pendulum: I*(d2theta/dt2)=-mgl*sin theta
period (T) of a simple pendulum: T=2pi*sqrt(l/g)

I see this system as essentially a simple pendulum. The mass of the wheel seems irrelevant, because it is distributed uniformly in such a way that it cannot affect the oscillation. The first formula above for rotational inertia is the only one I know for a simple pendulum that includes mass, the quantity we are looking for (the mass of the valve stem). But there are too many unknowns to solve for the mass.

The second formula for period should apply to this problem, but the given information is not consistent with the equation. Namely 2pi*sqrt(.3/9.81)=1.1 s, not 12s as given in the problem. Your thoughts?

 

[kuruman]

I see this system as essentially a simple pendulum.

More correctly a physical pendulum. What is the equation for the period of a physical pendulum?

 

[PeroK]

Homework Statement: A bicycle is turned upside down and the front wheel executes a slow, small-amplitude back-and-forth rotational motion with a period of 12 s. Considering the wheel to be a thin ring of mass 600 g and radius 30 cm, whose only irregularity is the presence of the tire valve stem, determine the mass of the valve stem.
Relevant Equations: rotational inertia (I) for simple pendulum: I*(d2theta/dt2)=-mgl*sin theta
period (T) of a simple pendulum: T=2pi*sqrt(l/g)

I see this system as essentially a simple pendulum. The mass of the wheel seems irrelevant, because it is distributed uniformly in such a way that it cannot affect the oscillation. The first formula above for rotational inertia is the only one I know for a simple pendulum that includes mass, the quantity we are looking for (the mass of the valve stem). But there are too many unknowns to solve for the mass.

The second formula for period should apply to this problem, but the given information is not consistent with the equation. Namely 2pi*sqrt(.3/9.81)=1.1 s, not 12s as given in the problem. Your thoughts?

When the valve is moving, the wheel is moving. When the valve stops, the wheel stops. What's happened to the angular momentum of the wheel?

 

[Brian_D]

Thank you, kuruman. So here are my further attempts, which still don't get to the answer. The period (T) of a physical pendulum is T=2pi*sqrt(I/mgL). To solve this, I note that I (rotational inertia) equals m*L^2. Substituting this expression into the formula for period, we get T=2pi*sqrt(m*L^2/mgL)=2pi*sqrt(L/g), which is the same as the equation for a simple pendulum. That gets me back to where I started. It also raises the new question why there is a separate formula for period of a physical pendulum when that formula reduces to the same one as for period of a simple pendulum. P.S. The answer to this problem according to the book's answer key is 5.0 g.

 

[PeroK]

P.S. The answer to this problem according to the book's answer key is 5.0 g.

That's what I get.

 

[Orodruin]

As has been alluded to already, you cannot ignore the mass of the wheel itself, it is highly relevant as it adds additional moment of inertia to the system. In other words, the total moment of inertia is not $mL^2$.

 

[gmax137]

That's what I get.

Me too.

 

[Orodruin]

That's what I get.

I get 5.1 g ...
(Using 9.81 m/s^2 for $g$ as OP did)

Actually, scrap that. The valve mass is supposedly part of the 600 g.

 

[PeroK]

I'm not sure an answer more precise that 5g is justified. Especially given the valve itself is probably inside the rim. Moreover, in practice I doubt the valve would cause the wheel to rotate. Not on my bike anyway - there's too much friction at the hub.

 

[bob012345]

I'm not sure an answer more precise that 5g is justified. Especially given the valve itself is probably inside the rim. Moreover, in practice I doubt the valve would cause the wheel to rotate. Not on my bike anyway - there's too much friction at the hub.

Well, it’s a mathematical bike.

 

[Orodruin]

Well, it’s a mathematical bike.

It is a massless spherical bike in a vacuum.

 

[bob012345]

Thank you, kuruman. So here are my further attempts, which still don't get to the answer. The period (T) of a physical pendulum is T=2pi*sqrt(I/mgL). To solve this, I note that I (rotational inertia) equals m*L^2. Substituting this expression into the formula for period, we get T=2pi*sqrt(m*L^2/mgL)=2pi*sqrt(L/g), which is the same as the equation for a simple pendulum. That gets me back to where I started. It also raises the new question why there is a separate formula for period of a physical pendulum when that formula reduces to the same one as for period of a simple pendulum. P.S. The answer to this problem according to the book's answer key is 5.0 g.

Are you still stuck on this point? It helps to actually derive the formula from scratch instead of doing plug and chug. Use Newton’s Second Law for angular acceleration and the period falls right out.

 

[Brian_D]

Thank you, bob012345, but I still don't get this.

As has been alluded to already, you cannot ignore the mass of the wheel itself, it is highly relevant as it adds additional moment of inertia to the system. In other words, the total moment of inertia is not $mL^2$.

As has been alluded to already, you cannot ignore the mass of the wheel itself, it is highly relevant as it adds additional moment of inertia to the system. In other words, the total moment of inertia is not $mL^2$.

OK, then how do you determine the moment of inertia? My textbook (Wolfson's and Pasachoff's Physics, Third Edition) has a table of rotational inertias for various shapes (p. 296), and the one for a thin ring is $I=MR^2$. If this is not correct, what expression are you using and how do you account for the mass of the wheel in that expression separately from the mass of the valve stem?

As has been alluded to already, you cannot ignore the mass of the wheel itself, it is highly relevant as it adds additional moment of inertia to the system. In other words, the total moment of inertia is not $mL^2$.

 

[Brian_D]

Are you still stuck on this point? It helps to actually derive the formula from scratch instead of doing plug and chug. Use Newton’s Second Law for angular acceleration and the period falls right out.

I see that torque is a form of Newton's Second Law, but don't see why rotational inertia is not $mR^2$, as I just explained in my reply to Orodruin.

 

[bob012345]

I think that comment meant it’s not little $m$ meaning just the valve. Your $MR^2$ is correct. Now consider what the torque is. Only the mass of the valve contributes to the torque because the rest of the wheel mass cancels out due to symmetry. But the whole wheel mass contributes to the inertia.

 

[Brian_D]

When the valve is moving, the wheel is moving. When the valve stops, the wheel stops. What's happened to the angular momentum of the wheel?

Thank you, but I don't see where you're going with this. The motion stops because of friction, right? But I think the problem is assuming a frictionless system.

 

[Brian_D]

I think that comment meant it’s not little $m$ meaning just the valve. Your $MR^2$ is correct. Now consider what the torque is. Only the mass of the valve contributes to the torque because the rest of the wheel mass cancels out due to symmetry. But the whole wheel mass contributes to the inertia.

Eureka! Thank you bob012345, this explanation of yours nailed it for me. Now I understand the calculation and also get 5.0 g.

 

[bob012345]

Eureka! Thank you bob012345, this explanation of yours nailed it for me. Now I understand the calculation and also get 5.0 g.

Thanks. Don’t feel bad, I’ve been doing these kinds of problems for about a half century and I still had to think about this one for a few minutes.