Bicycle wheel forced to roll around a vertical axle

[Nexus99]

Homework Statement

A wheel of a bike, equiparable to a ring of mass m and radius r, is connected through an orizzontal axis OO' to a vertical axis. The axis OO' mantain the wheel in vertical position: its lenght is d and its mass is negligible. An engine at t=0, transmit to the vertical axis a moment that increase following the law $M = 2t$, so the wheel, at the beginning, rolls without slipping. Calculate for t = 10 s, the angular momentum of the wheel respect to the point O' and the kinetic energy of the system. Calculate the moment when the wheel starts slipping. The static friction coefficient is $\mu_s$

Relevant Equations

angular momentum, momentum, pure rolling condition

I had some idea to solve this problem but i can't understand where the moment M is directed and where the force/forces that has magnitude $\frac{M}{d}$ is/are directed. Can anyone help me?

 

[etotheipi]

I believe the torque is applied by the rotating pole to the rod/wheel system. That torque only has a vertical ($\hat{z}$) component, so contributes to the rate of increase in angular momentum $L_z = md^2 \omega_z$ in the vertical direction.

If you take the origin of coordinates at the position where the rod meets the pole, then the friction force from the ground, $f_s \leq \mu N$, results in a torque which has a radial ($\hat{r}$) component as well as a vertical ($\hat{z}$) component. (N.B. this has been edited and corrected, after a mistake was pointed out by haruspex. Thanks!)

That should allow you to write two equations, one equation in terms of the angular acceleration $\alpha_z$ of the configuration around the pole, and another equation in terms of the angular acceleration $\alpha_r$ of the wheel about its centre.

You can finally apply the rolling condition, and solve for the critical case of the inequality $f_s \leq \mu N$ in terms of known variables.

 

[haruspex]

I believe the torque is applied by the rotating pole to the rod/wheel system. That torque only has a vertical ($\hat{z}$) component, so correlates to the rate of increase in angular momentum $L_z = md^2 \omega_z$ in the vertical direction. The friction force from the ground, $f_s \leq \mu N$, results in a torque about the centre of the wheel which has only a radial ($\hat{r}$) component, so correlates to the rate of increase in angular momentum $L_r = mr^2 \omega_r$ in the radial direction (assuming the wheel is a "hoop").

That should allow you to write two equations, one equation in terms of the angular acceleration $\alpha_z$ of the configuration around the pole, and another equation in terms of the angular acceleration $\alpha_r$ of the wheel about its centre.

You can finally apply the rolling condition, and solve for the critical case of the inequality $f_s \leq \mu N$ in terms of known variables.

The friction also produces a torque around the vertical axis, opposing the torque from the motor.

 

[etotheipi]

The friction also produces a torque around the vertical axis, opposing the torque from the motor.

That's right, thanks for pointing out, I missed this. I updated my solution

 

[haruspex]

where the force/forces that has magnitude Md is/are directed.

Why do you mention a force of magnitude M/d? Is there some piece of information you have omitted? Or are you just figuring that M equates to a force M/d exerted at distance d from O'?

 

[Nexus99]

Why do you mention a force of magnitude M/d? Is there some piece of information you have omitted? Or are you just figuring that M equates to a force M/d exerted at distance d from O'?

No, $M$ is a torque, so I thought $\frac{M}{d}$ was the force applied to the rod (or to the cm of the wheel and in opposite direction of the friction)

 

[haruspex]

No, $M$ is a torque, so I thought $\frac{M}{d}$ was the force applied to the rod (or to the cm of the wheel and in opposite direction of the friction)

Ok, that is the second option I mentioned,

just figuring that M equates to a force M/d exerted at distance d from O'

Yes, for the purposes of this question, you can just treat the motor as applying a horizontal force M/d through the centre of the wheel and in the plane of the wheel.
(More generally, you might also have to consider the centripetal force from the tension in the rod.)
So now you just have a wheel with a few forces on it. What are they and what equations can you write?

 

[Nexus99]

$\frac{M}{d} = F$
$f =$ friction
projecting the forces on the transverse direction to the circumference described by the wheel around the vertical axis:
$F - f = m a$

Torque about z axis, perpendicular to the plane of the wheel:
$fr = I \alpha$ $\rightarrow$ $f = ma$

So:
$a = \frac{t}{d}$

and deriving: $v = \frac{t^2}{2d}$

Now I'm not sure but i think that are present 2 angular velocity:
one is the angular velocity due to the rotation about the vertical axis: $\omega_1 = \frac{v}{d}$
the second the angular velocity due to the rotation about the point of contact of the wheel : $\omega_2 = \frac{v}{r}$

 

[haruspex]

$\frac{M}{d} = F$
$f =$ friction
projecting the forces on the transverse direction to the circumference described by the wheel around the vertical axis:
$F - f = m a$

Torque about z axis, perpendicular to the plane of the wheel:
$fr = I \alpha$ $\rightarrow$ $f = ma$

So:
$a = \frac{t}{d}$

How did you get that last line? It is dimensionally wrong.

 

[Nexus99]

How did you get that last line? It is dimensionally wrong.

I noticed it too. I only did algebra
$F = \frac{M}{d} = 2 ma$
And considering that $M = 2t$ you get my result. Probably the book forgot a coefficient in that law

 

[haruspex]

I noticed it too. I only did algebra
$F = \frac{M}{d} = 2 ma$
And considering that $M = 2t$ you get my result. Probably the book forgot a coefficient in that law

Ah, yes, I forgot you are given M=2t. It's ok to specify a relationship like that, but only if suitable units are prescribed. E.g. the 2 could be given as 2 Nm/s. Then you would have M=2t Nm/s, which is dimensionally ok.
But shouldn't you get $a=\frac t{md}$ Nm/s?

 

[Nexus99]

Ah, yes, I forgot you are given M=2t. It's ok to specify a relationship like that, but only if suitable units are prescribed. E.g. the 2 could be given as 2 Nm/s. Then you would have M=2t Nm/s, which is dimensionally ok.
But shouldn't you get $a=\frac t{md}$ Nm/s?

Yes, i forgot to write the mass, and yes, i saw now on the book that M = 2t Nm/s.

 

[Nexus99]

Ok, probably i finished the problem
$\omega_1 = \frac{v}{d}$
$\omega_2 = \frac{v}{r}$
$v = \frac{t^2}{2md}$

$L_{O'} = L' + L_{CM} = I_{CM} \omega_2 + \frac{I_{CM}}{2} \omega_1 + mvd$
where $I_{CM} = mr^2$

$K = K' + K_{CM}$

Last question can be find imposing that
$f ≤ \mu_s mg$

 

[haruspex]

$L_{O'} = L' + L_{CM} = I_{CM} \omega_2 + \frac{I_{CM}}{2} \omega_1 + mvd$
where $I_{CM} = mr^2$

It only makes sense to add up the terms like that if they are vectors, but you have not shown them as such.

 

[Nexus99]

So should be:
$L_{O'} = \sqrt{(I_{CM} \omega_2)^2 + (\frac{I_{CM}}{2} \omega_1 + mvd))^2}$

 

[haruspex]

So should be:
$L_{O'} = \sqrt{(I_{CM} \omega_2)^2 + (\frac{I_{CM}}{2} \omega_1 + mvd))^2}$

It says "calculate the moment", not the magnitude of it, so maybe you should leave it as a vector.

 

[Nexus99]

It says "calculate the moment", not the magnitude of it, so maybe you should leave it as a vector.

Ok so:
$\vec{L_{O'}} = I_{CM} \omega_2 \hat{z} - (\frac{I_{CM} \omega_1}{2} + m v d) \hat{y}$

 

[haruspex]

Ok so:
$\vec{L_{O'}} = I_{CM} \omega_2 \hat{z} - (\frac{I_{CM} \omega_1}{2} + m v d) \hat{y}$

How are you defining those axes? You seem to have $\hat y$ vertical, which is unusual.
Also, remember you are asked for the moment at 10s.

 

[Nexus99]

How are you defining those axes? You seem to have $\hat y$ vertical, which is unusual.
Also, remember you are asked for the moment at 10s.

y antiparallel to M and z perpendicular to the plane represented by the wheel

 

[haruspex]

y antiparallel to M and z perpendicular to the plane represented by the wheel

So that's a rotating frame?

 

[Nexus99]

So that's a rotating frame?

Yes but the frame rotates around y axis