Rolling with slipping and combined momentums

In summary: You can't add/compare two quantities with different units. The linear momentum and angular momentum of a rigid body are different things entirely.
  • #1
looooop
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Summary:: combining angular and linear momentum when an impulse is aplied 2/3 of the radius from the center.

A Jo-jo is lying on the ground on its edge. The central part (axel) has a radius of 2r and it’s side a radius of 3r. The string is protruding from the bottom of the axel (central part). The inertia (I) of the jo-jo is 15Mr^2.

Question one: a sudden force (an impulse) is applied to the string yanking it. This force is much grater then the friction so friction may be disregarded. After the impulse has ended the jo-jo has a angular speed of w and a linear speed of v. How big was the impulse?
SO I’ve tried to look at the momentum (mv) and angular momentum (Iw). But since the impulse imparts both rotation and linear velocity I’m unsure how to solve it. My thought is that since the impulse is aplied 2/3 of the way out from the center. 2/3 of the impluse goes towards angular momentum and 1/3 towards linear momentum. But I'm not really sure how to show this or if it's entirely correct

Question two: After the impulse has ended how long does it take for the jo-jo to start slipping and start roling with a dynamic friction koeficient og u. my though is to look at the torque generated by the rotasjon and the somhow add the linear movment to this. When this is equal to the friction force the jo-jo wil stop sliding and start roling. I'm unsure how to add the linear and rotasjonal parts or if it's correct and how i would match this to the friction?
 
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  • #2
looooop said:
My thought is that since the impulse is aplied 2/3 of the way out from the center. 2/3 of the impluse goes towards angular momentum and 1/3 towards linear momentum.

Angular momentum and linear momentum have different units and are different quantities so it doesn't make sense to compare them; i.e. there is no splitting.

You can consider them independently. The string applies a force to the jo-jo. The impulse of this force equals the change in momentum. The angular impulse of this force equals the change in angular momentum. The same force can be thought of as causing both effects independently of each other.

If the angular impulse is ##\tau \Delta t##, can you relate the linear impulse ##I = F\Delta t## to the angular impulse?
 
  • #3
etotheipi said:
Angular momentum and linear momentum have different units and are different quantities so it doesn't make sense to compare them; i.e. there is no splitting.

You can consider them independently. The string applies a force to the jo-jo. The impulse of this force equals the change in momentum. The angular impulse of this force equals the change in angular momentum. The same force can be thought of as causing both effects independently of each other.

If the angular impulse is ##\tau \Delta t##, can you relate the linear impulse ##I = F\Delta t## to the angular impulse?

I thought I could combine them since angular momentum = radius X mv. is this not possible?

Should it be impulse = momentum equation 1
Impulse = Angular momentum equation2
 
  • #4
looooop said:
I thought I could combine them since angular momentum = radius X mv. is this not possible?

You can't add/compare two quantities with different units. The linear momentum and angular momentum of a rigid body are different things entirely.

The angular momentum is not the radius ##\times## mv here. You're thinking of the formula ##L = mvr\sin{\theta}## which applies to a point particle moving with speed ##v## with respect to the origin of a coordinate system (not the speed of the origin itself!).

In this example, it is most convenient to take torques about the centre of mass of the body. The angular momentum about the axis through the CM is ##I\omega##, and the change in angular momentum about this axis equals the angular impulse about this axis.

The key thing is that a single force ##F## causes both an angular impulse and a linear impulse. Treat them separately!
 
  • #5
Thank you
 
  • #6
I'm still struggeling a bit here. The impulse is the only external force applied. it imparts both rotasjon and linear movment. So when trying to define how large the impulse is it seems to me that i have to define it using both the angular and linear veloceties. if i treat them separate i get an angular momentum Iw of 15MR^2w with the units kg m^2/s which doesn't match the units of the impulse. I'm thinking maybe divide by the radius at which the impulse is implied? but if you say they can't be combined is the anwser really just that the impulse is equall to the linear momentum? If so how do we acount for rotasjon imparted?
 
  • #7
looooop said:
The impulse is the only external force applied.

The impulse is not a force, it is a product of the force and an internal of time. I will give you the full explanation just so there is less change of misconceptions creeping in.

The resultant force on a rigid body ##\vec{F} = \frac{d\vec{P}}{dt}##, where ##\vec{P}## is the momentum of the rigid body. You can derive (or look at Euler's laws of motion) that the momentum of a rigid body is the total mass multiplied by the velocity of the centre-of-mass, i.e. ##\vec{P} = M\vec{V}_{CM}##. Hence we can say that ##\vec{F} = M\vec{A}_{CM}##.

Now if a resultant force ##\vec{F}## acts on the body between times ##t_1## and ##t_2##, then the impulse can be obtained via integrating Newton's second law: $$\vec{F} = \frac{d\vec{P}}{dt} \implies \int_{t_1}^{t_2} \vec{F} dt = \int_{\vec{P}_1}^{\vec{P}_2} d\vec{P} = \Delta (M\vec{V}_{CM})$$ If the force is constant, this just reduces to ##\vec{F}\Delta t = M\Delta \vec{V}_{CM}##. That is your first equation.

Now we do the same thing using torques, with the aim of obtaining an equation pertaining to the angular impulse of this force. We use $$\vec{\tau} = \frac{d\vec{L}}{dt} \implies \int_{t_1}^{t_2} \vec{\tau} dt = \int_{\vec{L}_1}^{\vec{L}_2} d\vec{L}$$If we only consider rotation about one axis coming out of the page (i.e. just take the ##z## component of this vector equation), and also assume the torque is constant, we can write this as just $$\tau_z \Delta t = \Delta L_z = \Delta (I_z \omega_z)$$ But what is ##\tau_z##? Well, in this case since the force is orthogonal to the radial vector, ##\tau_z = Fr##. So the angular impulse, which equals the change in angular momentum, equals ##Fr\Delta t##. But ##F\Delta t## is the linear impulse in this case!

So we conclude that in this case the angular impulse (which equals the change in angular momentum) is just the linear impulse (which equals the change in linear momentum) multiplied by the radius ##r##.

Notice that we treated the angular and linear effects of the same force separately, and now have two different equations pertaining to the angular and linear impulses. Hope this helps!
 
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  • #8
Wouldn't this then mean that the total impulse imparted is J = mv + mvr ?

to put it like this if i have one jo-jo like described and i apply F for dt to the string as a linear impulse. The impulse when it is transferred to the jo-jo from the string wil have a angular and a linear component.
 
  • #9
looooop said:
Wouldn't this then mean that the total impulse imparted is J = mv + mvr ?

to put it like this if i have one jo-jo like described and i apply F for dt to the string as a linear impulse. The impulse when it is transferred to the jo-jo from the string wil have a angular and a linear component.

No. The linear impulse is ##F\Delta t = \Delta P = Mv_f##, where ##v_f## is the final velocity of the centre of mass. The angular impulse is ##Fr \Delta t = I\omega_f## where ##\omega_f## is the final angular velocity. I suppose you could say in this case that the angular impulse is ##Mrv_f##, though I'm not sure if that's helpful.

But it makes no sense to add them together. For starters, they have different units!

I think the wording of the question is perhaps misleading. When it asks how large the impulse is, I assume it means the linear impulse. For this you don't need to know anything about the angular velocity or the moment of inertia. That's only important for the second part.
 
  • #10
looooop said:
The central part (axel) has a radius of 2r and it’s side a radius of 3r. The string is protruding from the bottom of the axel (central part). The inertia (I) of the jo-jo is 15Mr^2.
Where M is the total mass? It cannot be more than 9Mr2. Are you sure you have quoted this correctly?
looooop said:
After the impulse has ended the jo-jo has a angular speed of w and a linear speed of v. How big was the impulse?
Strange... too much information. You only need the mass and the speed, or, the angular speed, MoI and point of application of the impulse.
looooop said:
for the jo-jo to start slipping and start roling
Stop slipping.
 
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  • #11
woops yes stop slipping of course.

M is a unit value of mass 5kg or whatever
 
  • #12
looooop said:
M is a unit value of mass 5kg or whatever
Ah, so we are not told the mass of the yoyo. That explains why we are told v and ω.
etotheipi said:
When it asks how large the impulse is, I assume it means the linear impulse. For this you don't need to know anything about the angular velocity or the moment of inertia.
No, since it turns out we do not know the mass, angular velocity and the moment of inertia is the way to solve the first part .
 
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  • #13
haruspex said:
No, since it turns out we do not know the mass, angular velocity and the moment of inertia is the way to solve the first part .

I see. I hadn't looked carefully at the MoI and had just assumed we knew the mass to be ##M##. Sorry!
 
  • #14
haruspex said:
Ah, so we are not told the mass of the yoyo. That explains why we are told v and ω.

No, since it turns out we do not know the mass, angular velocity and the moment of inertia is the way to solve the first part .

we know the total mass to be 5M. Does this change anything in regards to combining angular and linear impuls? It was my understanding that it was an aspect of their properties that they coulden't be combined
 
  • #15
looooop said:
we know the total mass to be 5M. Does this change anything in regards to combining angular and linear impuls? It was my understanding that it was an aspect of their properties that they coulden't be combined

It has no effect on the latter sentence; they are distinct quantities and can't be combined.

But if you know the mass to be 5M (was this omitted from what you included of the problem statement?) now you have two ways of calculating the linear impulse :wink:.
 

Related to Rolling with slipping and combined momentums

1. What is rolling with slipping and combined momentums?

Rolling with slipping and combined momentums is a concept in physics that describes the motion of an object that is both rolling and slipping at the same time. It takes into account the rotational and translational velocities of the object, as well as the forces acting on it.

2. How is rolling with slipping different from pure rolling?

Rolling with slipping occurs when an object is rolling and slipping at the same time, while pure rolling is when an object is rolling without any slipping. In pure rolling, the point of contact between the object and the surface it is rolling on is stationary, while in rolling with slipping, the point of contact is moving.

3. What factors affect the combined momentum of an object rolling with slipping?

The combined momentum of an object rolling with slipping is affected by several factors, including the mass and velocity of the object, the friction between the object and the surface, and the shape and size of the object.

4. How does the friction force affect rolling with slipping?

Friction is an important factor in rolling with slipping, as it is the force that causes the slipping motion. The amount of friction between the object and the surface will determine the amount of slipping and therefore, the combined momentum of the object.

5. Can rolling with slipping be used to model real-world situations?

Yes, rolling with slipping can be used to model real-world situations, such as the motion of a car's tires on a road or a ball rolling down a hill. It is a useful concept in understanding and predicting the motion of objects in various scenarios.

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