How can we model a rolling truck on an incline using rotational motion analysis?

[vcsharp2003]

Homework Statement

A truck rolls down a 7.8 degree hill with a constant speed of 30.0 m/s. At the bottom of the hill it continues on a horizontal surface. How far will the truck go before it stops?

Relevant Equations

F = ma

This is a problem that was posted here in 2003 and is now closed for replies. This question can be found at https://www.physicsforums.com/threads/friction-problem.662/

The answer in that old post didn't seem clear to me probably because it was highly summarized. There was no mention of static friction which acts in the direction of motion for a rolling object with no slippage and also of rolling friction that acts opposite to the direction of motion of center of the rolling object.

Based on my analysis, I came up the with following free body diagram of the truck wheels while on incline and also on horizontal ground. I'm not sure if the frictional forces shown are correct for this scenario. $f_r$ denotes the rolling friction force from inclined ground on the truck wheels and $f_s$ the static frictional force from horizontal ground on the truck wheels.

For truck wheels rolling on incline:
$mg sin {\theta} + f_s = f_r$ and $R_1 = mg cos {\theta}$

For truck wheels rolling on horizontal ground:
$f_s - f_r = -ma$ and $R_2 = mg$

We know $\theta = 7.8^o$ and when the truck is moving on horizontal ground then its experiencing a deceleration $a$ that is pointing towards left. $a$ denotes the magnitude of deceleration.

I have probably committed a mistake in my analysis since I have assumed rolling friction forces to be equal on inclined and horizontal surfaces, and also I have assumed that static friction forces to be equal on both surfaces, which I think is not necessarily true. In that case, it appears that this problem is not solvable with the given information.

 

[vcsharp2003]

Homework Statement:: A truck rolls down a 7.8 degree hill with a constant speed of 30.0 m/s. At the bottom of the hill it continues on a horizontal surface. How far will the truck go before it stops?
Relevant Equations:: F = ma

In that case, it appears that this problem is not solvable with the given information.

I guess, even equations of rotational motion about the point of contact of truck wheels with ground will need to be considered to solve this problem. Then, it will be solvable.

$\tau = I \alpha$
$v = r \omega$
$a = r \alpha$

 

[haruspex]

Homework Statement:: A truck rolls down a 7.8 degree hill with a constant speed of 30.0 m/s. At the bottom of the hill it continues on a horizontal surface. How far will the truck go before it stops?
Relevant Equations:: F = ma

static friction which acts in the direction of motion for a rolling object

Not necessarily. The motion of a free-rolling truck downhill is opposed by rolling resistance, drag and axle friction, while gravity tends to accelerate it.
Since the wheels are rotating at a constant rate, there is no net torque on them. What directions of torque arise from these various forces?

 

[vcsharp2003]

Not necessarily.

You mean static friction force could be in a direction opposite to direction of velocity of center? If yes, then what is the reasoning behind this.

For a car on a road the static friction acts in the direction of car's velocity, and the reason is that friction supplies the forward force to accelerate the car.

 

[haruspex]

For a car on a road the static friction acts in the direction of car's velocity, and the reason is that friction supplies the forward force to accelerate the car.

The answer is partly in what you wrote: to accelerate the car. What if the car is at constant velocity, or constant speed on a bend, or slowing down?
Also need to consider gradient. Consider a car going downhill, gaining speed, but more slowly than if freewheeling.

 

[vcsharp2003]

The answer is partly in what you wrote: to accelerate the car. What if the car is at constant velocity, or constant speed on a bend, or slowing down?

So, the static friction force must be in a direction to support it's resultant motion.
The direction of static friction then is not fixed for rolling object, which is what I was thinking up until now.
If the car is at a constant speed on a straight road then static friction force must be 0 as per Newton's second law of motion.
If the car is turning along a circular arched road at a constant speed, then the static friction force must be acting towards the center of circular arc so it can supply the necessary centripetal force.

It makes sense and thankyou for opening my eyes to this fact.

 

[vcsharp2003]

Since the wheels are rotating at a constant rate, there is no net torque on them. What directions of torque arise from these various forces?

I get the following net torque about points P and O. When taking torque about point P, the net torque cannot be zero, but about point O it seems possible. I don't know why this is happening? Do I need to change my free body diagram for object rolling on the inclined surface?

 

[vcsharp2003]

Do I need to change my free body diagram for object rolling on the inclined surface?

Probably show $f_s$ as acting up the incline at point P. If yes, then why would this be? Is it because rolling friction is always very small and therefore, cannot balance the larger $mg \sin {\theta}$ force?

Probably, $f_s>>>f_r$.

 

[haruspex]

Probably show $f_s$ as acting up the incline at point P. If yes, then why would this be? Is it because rolling friction is always very small and therefore, cannot balance the larger $mg \sin {\theta}$ force?

Probably, $f_s>>>f_r$.

I assume what you are calling rolling friction is what I prefer to call rolling resistance.
It is a torque rather than a force. It arises from a shift in the application profile of the normal force. See section 4.1 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/.
If the car is braking or has axle friction, that's another torque opposing motion.
Any aerodynamic drag on the car will lead to a horizontal component to the load it places on the axles.

 

[vcsharp2003]

I assume what you are calling rolling friction is what I prefer to call rolling resistance.

So, is it correct to assume $f_s$ as opposing translational motion of wheel on incline? And the reason for it should be that $f_s>>>f_r$ and the weight component along incline could never be balanced for constant velocity motion of wheel by such a small rolling friction force.

 

[haruspex]

So, is it correct to assume $f_s$ as opposing translational motion of wheel on incline? And the reason for it should be that $f_s>>>f_r$ and the weight component along incline could never be balanced for constant velocity motion of wheel by such a small rolling friction force.

As I posted, Rolling resistance is a torque, not a force, so the comparison is between that ($\tau_r$ say) and $Rf_s$, where R is the wheel's radius.
If you deflate the tyres, and maybe part fill them with sand, you could get considerable rolling resistance.
The only way I can see that the static friction could act in the direction of motion is if the engine is engaged and, together with gravitational assistance, only just balancing the combination of drag and rolling resistance. That could happen on a gentle incline.

 

[vcsharp2003]

I get the following net torque about points P and O. When taking torque about point P, the net torque cannot be zero, but about point O it seems possible. I don't know why this is happening? Do I need to change my free body diagram for object rolling on the inclined surface?

View attachment 301131

Why is torque about point P never becoming zero in above diagram? If we consider the whole truck as single system with rolling wheels, then the axle friction would be an internal force to the system. Then, the net torque about P is never going to be zero on an incline, whereas net torque about any point must be zero if truck is moving at constant velocity.

 

[kuruman]

Homework Statement:: A truck rolls down a 7.8 degree hill with a constant speed of 30.0 m/s. At the bottom of the hill it continues on a horizontal surface. How far will the truck go before it stops?
Relevant Equations:: F = ma

The answer in that old post didn't seem clear to me probably because it was highly summarized. There was no mention of static friction which acts in the direction of motion for a rolling object with no slippage and also of rolling friction that acts opposite to the direction of motion of center of the rolling object.

I think you are overthinking this and got bogged down in the modeling details. For the purposes of answering the question posed, it suffices to assume that the effect of the non-conservative forces on the motion, whatever they are (rolling friction, air resistance, the driver applying the brakes, etc.), does not change when the truck moves from the inclined to the horizontal part of its path. Then one can use the work-energy theorem with the rates at which the conservative forces (gravity) and the non-conservative forces do work written separately.

For the inclined part there is no change in kinetic energy so that $$0=\frac{dW_{\text{grav.}}}{dt}+\frac{dW_{\text{N.C.}}}{dt}\implies \frac{dW_{\text{N.C.}}}{dt}=-\frac{d(mgh)}{dt}.$$One can easily find an expression for the rate at which gravity does work on the truck and evaluate it numerically in terms of the given quantities (the mass of the truck will drop out in the end). Then for the flat part where the kinetic energy is changing but gravity does no work, $$\frac{dK}{dt}=\frac{dW_{\text{N.C.}}}{dt}= -\frac{d(mgh)}{dt}.$$One can use this last expression to find the horizontal distance traveled in any number of ways.

You did not provide a link to the old post and one cannot see how the question was answered there, but I would do it as outlined above. It's OK to worry about the details of the retarding forces and free body diagrams if you find it instructive to do so. However such details, when handled properly, will not give a different answer to the question posed.

 

[vcsharp2003]

You did not provide a link to the old post and one cannot see how the question was answered there, but I would do it as outlined above

I've provided the link in my original post. Its https://www.physicsforums.com/threads/friction-problem.662/

 

[jbriggs444]

I think you are overthinking this and got bogged down in the modeling details. For the purposes of answering the question posed, it suffices to assume that the effect of the non-conservative forces on the motion, whatever they are (rolling friction, air resistance, the driver applying the brakes, etc.), does not change when the truck moves from the inclined to the horizontal part of its path. Then one can use the work-energy theorem with the rates at which the conservative forces (gravity) and the non-conservative forces do work written separately.

For the inclined part there is no change in kinetic energy so that $$0=\frac{dW_{\text{grav.}}}{dt}+\frac{dW_{\text{N.C.}}}{dt}\implies \frac{dW_{\text{N.C.}}}{dt}=-\frac{d(mgh)}{dt}.$$One can easily find an expression for the rate at which gravity does work on the truck and evaluate it numerically in terms of the given quantities (the mass of the truck will drop out in the end). Then for the flat part where the kinetic energy is changing but gravity does no work, $$\frac{dK}{dt}=\frac{dW_{\text{N.C.}}}{dt}= -\frac{d(mgh)}{dt}.$$One can use this last expression to find the horizontal distance traveled in any number of ways.

I agree that this is the intended approach. However, it depends on two idealizations.

1. As you already noted, the effect of air resistance must either be neglected or assumed to be constant throughout the deceleration phase.

2. The joining of the descending ramp and the horizontal track must either be rounded off so that no kinetic energy is lost at the junction or any such kinetic energy loss must be neglected. Alternately, the enterprising student can opt to conserve horizontal momentum through the partially elastic collision there.

 

[vcsharp2003]

Homework Statement:: A truck rolls down a 7.8 degree hill with a constant speed of 30.0 m/s. At the bottom of the hill it continues on a horizontal surface. How far will the truck go before it stops?
Relevant Equations:: F = ma

In that case, it appears that this problem is not solvable with the given information.

Also, I see a contradiction because of constant velocity of truck down the incline, when I apply law of conservation of energy.

If frictional losses are negligible, then energy at two instants that are a height $h$ apart on the incline plane can be equated like this: $mgh + \frac {1} {2} m {v_{cm}}^2 + \frac {1} {2} I {\omega}^2 = \frac {1} {2} m {v_{cm}}^2 + \frac {1} {2} I {\omega}^2$. Since center of mass of truck is moving with constant velocity $v_{cm}$ on the incline and also angular speed of each of the wheels $\omega$ is constant, so the energy is not being conserved.

Its as if truck must move with some acceleration down the incline else the energy equation gets violated.

 

[jbriggs444]

so the energy is not being conserved.

Of course, mechanical energy is not being conserved. That is the point of the problem. You are expected to pay attention to frictional losses, not ignore them.

Also, you are expected to neglect the rotational kinetic energy of the wheels, otherwise the answer can vary. You need the energy of the cart (per unit mass) to be a knowable function of velocity such as $KE=\frac{1}{2}mv^2$. Having it depend on wheel geometry as well ($KE=\frac{1}{2}mv^2 + \frac{1}{2}I \omega^2$) leaves you with too many unknowns.

 

[vcsharp2003]

One can use this last expression to find the horizontal distance traveled in any number of ways.

$$ \frac {d{(\frac {1} {2} m {v_{cm}}^2})} {dt} = -mg \frac {h} {dt}$$
$$\frac {1} {2} \times m \times 2v_{cm} \times \frac {dv_{cm}} {dt} = -mg \frac {d(s \times sin\theta)} {dt} $$
$$\frac {1} {2} \times m \times 2v_{cm} \times \frac {dv_{cm}} {dt} = -mg \times sin\theta \times \frac {d(s)} {dt} $$
$$\frac {1} {2} \times m \times 2v_{cm} \times \frac {dv_{cm}} {dt} = -mg \times sin\theta \times v_{cmi} $$
In above equation $v_{cm}$ is the center of mass speed on horizontal surface at an instant $t$, whereas $v_{cmi}$ is the initial speed of center of mass on horizontal surface. Then, we get
$$\therefore \frac {dv_{cm}} {dt} = -\frac {v_{cmi}gsin\theta} {v_{cm}}$$

I get a another value of deceleration using my original equations I had in my original post as shown below, but using the same assumption as what you used. I had also assumed that all frictional forces are identical for both the incline and horizontal surfaces. If I subtract equation#3 from equation#1 then I also get $a = gsin\theta$ as shown below. In my analysis $a$ was the magnitude of deceleration and hence I am getting a positive value unlike the above approach.

$$mg sin {\theta} + f_s - f_s + f_r= f_r +ma$$
$$\therefore a = g sin \theta$$

 

[vcsharp2003]

Of course, mechanical energy is not being conserved. That is the point of the problem. You are expected to pay attention to frictional losses, not ignore them.

Also, you are expected to neglect the rotational kinetic energy of the wheels, otherwise the answer can vary. You need the energy of the cart (per unit mass) to be a knowable function of velocity such as $KE=\frac{1}{2}mv^2$. Having it depend on wheel geometry as well ($KE=\frac{1}{2}mv^2 + \frac{1}{2}I \omega^2$) leaves you with too many unknowns.

By friction, you mean the air resistance, since friction at point of contact of wheels with ground does no work as point of contact has a zero speed. Or you meant some other dissipative forces?

 

[kuruman]

Alternately, the enterprising student can opt to conserve horizontal momentum through the partially elastic collision there.

I don't see how horizontal momentum is conserved. It is $mv\cos\theta$ until it becomes $mv$ under the assumption of no energy loss when the truck turns the corner. I guess, I am not enterprising, but then again I am not a student either.

 

[jbriggs444]

I don't see how horizontal momentum is conserved. It is $mv\cos\theta$ until it becomes $mv$ under the assumption of no energy loss when the truck turns the corner. I guess, I am not enterprising, but then again I am not a student either.

If one discards the [counter-physical] assumption of no energy loss at a sharp collision event then conservation of horizontal momentum remains somewhat plausible.

 

[jbriggs444]

By friction, you mean the air resistance, since friction at point of contact of wheels with ground does no work as point of contact has a zero speed. Or you meant some other dissipative forces?

The force of friction together with the velocity of the center of mass quantifies the rate at which mechanical energy is drained from the cart as a whole, even though no "work" is being done by tangential friction across the wheel/surface interface when you look at just that interface.

A term sometimes used for frictional force times displacement of the center of mass is "center of mass work".

It matters which velocity you dot with force.

 

[haruspex]

Why is torque about point P never becoming zero in above diagram?

Because you are only looking at one wheel axle. Whichever you look at, front or back, the mass centre of the car and the normal force from the ground at the other axle have torques about the point of contact of wheel on ground. The distribution of load between the normal forces can produce different torques for the same total force.

 

[Lnewqban]

While rolling downhill, the gravity driving force is perfectly balanced by the resistive forces, since there is no acceleration.
You could replace the shown situation with a truck rolling horizontally, being the exactly same driving force coming from its engine in this case.

In my view, the question of the problem is how long the truck would keep rolling on that horizontal surface after the driving force is suddenly removed from the picture, while the resistive (non-conservative) forces remain the same.

 

[vcsharp2003]

Because you are only looking at one wheel axle. Whichever you look at, front or back, the mass centre of the car and the normal force from the ground at the other axle have torques about the point of contact of wheel on ground. The distribution of load between the normal forces can produce different torques for the same total force.

You're right. Thanks once again for such a precise explanation.

After all these posts, my view is that if I'm looking at a rolling object on an incline as the model to solve this problem, then I must not consider the whole truck/car + 4 wheels system, since its a complex model with a 3 dimensional array of forces to be looked at and not really a rolling object; rather, I must look at a single wheel as the system to model in this problem and then apply the laws of rotation to this system. The wheel mimics a rolling object perfectly and therefore, we can easily apply rotational logic to it. The whole truck/car cannot mimic a rolling object and makes its analysis more complex/difficult.

If I do that then the net torque must be zero about any point in space and also the forces due to brakes or axle or by anything attached to the wheel can now be considered all external forces. If I took the whole car as the system to model then brake forces or axle forces on the wheel would be considered internal forces and therefore, not useful for analysis. These forces would now also contribute to a balancing torque when taking torque about point P and make the net torque zero explaining my doubt in post#12.

Taking the wheel as the system also explains the law of conservation of energy which seemed to be getting violated as per my post#16. Now, frictional work done by resistive forces to the rotating wheel system through some connected mechanism in truck/car would be doing negative work (because frictional torque would oppose angular displacement of wheel); essentially, the initial GPE in post#16 would be getting converted to frictional work and the conservation of energy equation would be clearly satisfied (frictional work magnitude would then come on the RHS of the energy equation in my post#16)

In short, by taking a single wheel as the system, all doubts get removed and it becomes a clean application of rotational motion.