Friction of a rolling cylinder on an incline

[malawi_glenn]

Does a static axis mean that some apparatus allows rotation about the center of mass but prevents translation?

Consider the front wheel on a bike as an example.

 

[PeroK]

Impulse is Newton seconds, velocity is meters per second, Energy is Newton meters, so I am guessing (but not sure) that you take the dot product of the impulse and velocity vector to get the work done.

Okay, but why more work for the same impulse at the rim compared to at the centre?

 

[Chenkel]

Okay, but why more work for the same impulse at the rim compared to at the centre?

I'm not sure how to answer that question, and I'm also not sure if my equation for energy is correct... I cannot keep treating objects as point masses, I'll only get so far with that I must learn nature's secret.

 

[PeroK]

I'm not sure how to answer that question, and I'm also not sure if my equation for energy is correct... I cannot keep treating objects as point masses, I'll only get so far with that I must learn nature's secret.

Try drawing a diagram of each case. Hint: don't make the time of the impulse too short.

 

[erobz]

I'm not trying to hijack Chenkels thread, but maybe someone could help us solve for the acceleration of the following wheel and it would be beneficial to both our understanding?

I'm imagining a constant torque applied to the wheel $T$. What is the acceleration of the wheel?

$$ \sum F_y = N - mg = 0 \implies N = mg \tag{1} $$

$$ \sum F_x = f_r = ma \tag{2}$$

From $(1)$:

$$ f_r = \mu mg $$

$\mu$ is not a constant with respect to $a$. I believe it must be the case that if the wheel is not slipping:

$$\mu = \frac{1}{g}a \leq \mu_s\tag{3}$$

Now applying a torque balance about the axis $\circlearrowright^+$

$$ \sum \tau = T - \mu mg R = I \alpha $$

With no slip : $\alpha = \frac{a}{R}$

$$T - \mu mg R =I \frac{a}{R} \tag{4}$$

Substitute $(3) \to (5)$:

$$ \implies a = \frac{T R}{ I + m R^2} $$

We have the condition that $a \leq \mu_s g$

Which means for the wheel not to slip:

$$ T \leq \mu_s g \frac{1}{R} \left( I + m r^2 \right)$$

correct?

EDIT:

It seems so be in agreement with what @haruspex gave in for the same geometry.

https://www.physicsforums.com/threads/a-spring-disk-and-pulley-system.1016701/post-6651212

$$ \tau \leq \frac{3}{2}\mu_s N R$$

 

[kuruman]

Which wheel is this? Can we have a picture?

 

[erobz]

Which wheel is this? Can we have a picture?

yeah, I forgot to put it in!

 

[kuruman]

I started a parallel thread here in the guise of a test to diagnose the doubts, confusion and misunderstandings that some participants may have and take corrective action. The idea is to keep this thread started by @Chenkel focused on the original question.