Engineering Rigid wheel rolling without slipping -- Trying to find angular acceleration

[Pipsqueakalchemist]

Homework Statement

Question below and attempt below

Relevant Equations

Newton’s law
Relative acceleration

So I tried the problem and it’s different from the solution. I’m confused on why my attempt didn’t work, is it because the wheel is undergoing general planar motion? I tried to just apply Newton’s 2nd law to find the acceleration of the centre and then use that to find angular acceleration. The solution uses relative acceleration and moment to solve the problem.

 

[PeroK]

I tried to just apply Newton’s 2nd law to find the acceleration of the centre and then use that to find angular acceleration. The solution uses relative acceleration and moment to solve the problem.

You have an unknown force of friction at the point of contact. You could try adding that force to your method.

 

[PeroK]

PS You should try to learn latex, as your solution is effectively unreadable. For example:
$$ma = F_h - f$$ $$I\alpha = \tau_h + \tau_f - \tau_v$$
Where I'll let you work out the notation I've used as a hint.

 

[Pipsqueakalchemist]

You have an unknown force of friction at the point of contact. You could try adding that force to your method.

Friction isn’t present in this problem. The FBD only has the normal force,weight, and the applied force

 

[Lnewqban]

Friction isn’t present in this problem. The FBD only has the normal force,weight, and the applied force

Friction force has an infinite value in this problem.
The ring-surface point of contact is actually considered a solid hinge about which all moments are calculated.
The solution does not show that force because it has no lever respect to that hinge, causing no moment or torque about it.

 

[PeroK]

Friction isn’t present in this problem. The FBD only has the normal force,weight, and the applied force

Well, then, the FBD is wrong! There's an anticlockwise torque on the ring from the applied force, so you'll need friction to get it rolling (clockwise) without slipping. It requires friction to initiate rolling without slipping.

 

[Pipsqueakalchemist]

Ok I see, but I thought friction isn’t present when it’s rolling without slipping. Isn’t it true that when rolling without slipping friction does no work? So if friction does no work then it doesn’t change the objects velocity so acceleration is equal to zero and its force is zero. That’s how I thought of it. And I’m also confuse about the direction of friction in this question. I see that it should be to the left to make the wheel spin clockwise but i thought that when you have a wheels and say it’s rolling to the right and an force is applied to the wheel that doesn’t pass through the centre of mass and creates an moment like in this question,I thought that friction is what moves the wheel forward so the friction force would be in the same direction of the wheels velocity.

 

[Pipsqueakalchemist]

Like in these two pictures, the one where friction acts to the left when the force passes through line of action. And I’m the other one friction is to the right when force doesn’t pass through centre of mass and creates an moment. In this question it’s the second case where force applied doesn’t pass through centre so friction should be directed to the right but I see that doesn’t make sense because if friction acts to the right the that's an counterclockwise direction.

 

[PeroK]

Ok I see, but I thought friction isn’t present when it’s rolling without slipping. Isn’t it true that when rolling without slipping friction does no work? So if friction does no work then it doesn’t change the objects velocity so acceleration is equal to zero and its force is zero. That’s how I thought of it. And I’m also confuse about the direction of friction in this question. I see that it should be to the left to make the wheel spin clockwise but i thought that when you have a wheels and say it’s rolling to the right and an force is applied to the wheel that doesn’t pass through the centre of mass and creates an moment like in this question,I thought that friction is what moves the wheel forward so the friction force would be in the same direction of the wheels velocity.

Rolling without slipping at constant velocity proceeds without friction. But, acceleration and deceleration require friction - it's hard to accelerate or brake on slippery ice, but not to keep moving.

In this case, the applied force tends to drag the ring across the surface. Friction, in the opposite direction, tends to reduce the acceleration and cause angular acceleration.

 

[PeroK]

Like in these two pictures, the one where friction acts to the left when the force passes through line of action. And I’m the other one friction is to the right when force doesn’t pass through centre of mass and creates an moment. In this question it’s the second case where force applied doesn’t pass through centre so friction should be directed to the right but I see that doesn’t make sense because if friction acts to the right the that's an counterclockwise direction.

In the first diagram, the force applied at the top of the ting tends to over-rotate it. The friction acts to counter this and to increase linear acceleration.

You need to analyse each scenario on its own merits, not jump to general conclusions about the direction of friction.

 

[PeroK]

PS In any case, you can always include a friction force without knowing its direction: balancing the equation will give you that.

 

[Pipsqueakalchemist]

In the first diagram, the force applied at the top of the ting tends to over-rotate it. The friction acts to counter this and to increase linear acceleration.

You need to analyse each scenario on its own merits, not jump to general conclusions about the direction of friction.

Ok I see, so my assumption about frictions were wrong

 

[PeroK]

Ok I see, so my assumption about frictions were wrong

You don't need to make assumptions. You have an equation for linear acceleration involving friction, an equation for angular acceleration also involving friction, and, for rolling without slipping, an equation relating linear to angular acceleration. Solving those equations gives you the unknown friction and the linear and angular accelerations for every scenario.

That was your alternative to the book answer in the original post.

 

[Pipsqueakalchemist]

But what about when rolling without slipping. I thought friction does no work when rolling without slipping and if it does no work then it doesn’t change the objects velocity meaning acceleration is zero so friction force equal zero because force = ma. Is there something wrong with my assumption?

 

[PeroK]

But what about when rolling without slipping. I thought friction does no work when rolling without slipping and if it does no work then it doesn’t change the objects velocity meaning acceleration is zero so friction force equal zero because force = ma. Is there something wrong with my assumption?

Do the calculations!

 

[PeroK]

But what about when rolling without slipping. I thought friction does no work when rolling without slipping and if it does no work then it doesn’t change the objects velocity meaning acceleration is zero so friction force equal zero because force = ma. Is there something wrong with my assumption?

Let's consider a ring being pulled by a horizontal force, $F$, through its centre of gravity a) on a smooth surface and b) on a rough surface, where we have an unknown frictional force $f$.

In case a) we have simply linear acceleration from $F = ma$ and no rotation.

In case b) we have linear acceleration given by $F - f = ma$ and an angular acceleration given by $fR = I\alpha$. And, of course, we have $a = R\alpha$.

We can, therefore, solve for $f$: $$f(1 + \frac{mR^2}{I}) = F$$ For a ring we have $I = mR^2$, hence $f = \frac{F}{2}$, which is definitely not zero.

In fact, it's clear that the frictional force is instrumental is causing some of the linear motion to become rotational motion - if we compare with case a).

The linear acceleration is: $$a = \frac{F}{2m}$$ Now we can check the work done. The linear kinetic energy after moving a distance $d$ is half what we would have in case a): $$KE_l = \frac{Fd}{2}$$ But, we also have rotational KE of $$KE_r = \frac 1 2 I \omega^2 = \frac 1 2 I \frac{v^2}{R^2} =\frac 1 2 mv^2 (\frac{I}{mR^2}) = KE_l$$ And the total kinetic energy (linear plus rotational) is the same as in case a).

In conclusion, the frictional force did no work. But, this absolutely didn't mean that it was zero.

If you do the same calculations for your more complicated problem, then you should get the book answer.