New answers tagged

0

I guess you need the moment of inertia of the car itself and not the wheels. Therefore you could determine this using the formula for a compound pendulum $T=2\pi \sqrt{\frac{I}{mgR}}$ and doing the experiment. For the experiment you need to hang your car from a rope or something similar and let it swing, measure the period $T$ and the radius $R$ to the ...


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Yes, the same approach can be used. The only difference is that for a car only part of the structure is rotating. When a ball is rolling down a ramp the whole ball is rotating so the moment of inertia you measure is the moment of inertia of the whole ball. For your car only the wheels are rotating, so the rotational energy you will be measuring is the ...


4

Since energy is conserved, the total kinetic energy after the yo-yo has fallen a distance $h$ is simply the initial energy, $E_1=mgh$. However, gravitational potential energy is converted into two forms of kinetic energy: translational kinetic energy ($\frac{1}{2}mv^2$) and rotational kinetic energy ($\frac{1}{2}I\omega^2$). Here, $I$ is the moment of ...


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Do you have to sum moments about points on an object, or can moments be summed around an object as well. In statics there are no restrictions about the point in space you choose, so that point need not be physically located on a point of the object for which you are determining static equilibrium. In fact, for an object to be in static equilibrium, the ...


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You can sum the moments around any point and get the correct results. Summing moments and summing forces are actually mathematically equivalent. You just have to integrate everything properly (basically summing over every particle). Some problems are easier to solve with moments, others are easier to solve with momentum. What's happening here is no more ...


0

Angular Acceleration is not the same as Translation Acceleration. Translational Acceleration in an object cannot come without any external force, but the angular acceleration in an object may come without any external torque. External torque is NOT the only source of angular acceleration. Angular acceleration in a rotating body may come even when the moment ...


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$$\pmb {\underline {\text {Short Answer}}}$$ The inertia frame cause of this is the inertia of the particle. $$\pmb {\underline {\text {Long Answer}}}$$ Suppose you are in an accelerating (in forward direction) train and you throw a ball upward. What happens? In real life it falls a away (backward) from the position it was expected to. You are in a bus ...


4

Rotating reference frame In the rotating reference frame the centrifugal force explains the sinking of dense particles, just as gravity does on the surface of earth. Unfortunately, the terminology “fictitious” makes it seem as though it cannot do anything. For that reason I prefer the term “inertial force” instead of “fictitious force”. In a non-inertial ...


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If you take a body with a given moment of inertia around some axis, then melt and recast it into a thin ring with the same mass, such that it has the same value of moment of inertia around the axis through its centre and perpendicular to the plane of the ring, as the original body, the radius of the ring that you get is the radius of gyration. If the moment ...


2

These off-diagonal terms are called the products of inertia. If we have an angular velocity vector $\boldsymbol{\omega}$, then the product of inertia $I_{ij}$ is the proportionality constant that measures how much the $j$th component of $\boldsymbol{\omega}$ contributes to the $i$th component of the angular momentum. This is a simple generalization of normal ...


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Consider the case 1. As I have shown the coefficient of limiting friction should be greater than or equal to the tan value of the angle of inclination.If not the equillibrium will be disturbed and the cone will slide down. Now lets assume that this condition is satisfied. Eventually as you keep on increasing the inclination we get to case 2. Once we ...


2

cylindrical rod of mass M and length l. Suppose that a force $\mathbf{F}$ of constant magnitude and direction (though movable) is applied on the rod at distance $r$ from the COM for a displacement $\boxed{d}$. I'm going to assume that $d$ is the displacement of the CM. Correct me if I'm wrong. I'll briefly describe the setup (so you can correct me if it's ...


2

If you apply the force a distance $r$ along the rod and during that time it rotates by a distance $d_{cm}$ and small_ angle $\theta$, then the point where you apply the force moves by $d_{cm} + r \theta$. This means the work done and energy added is $F d_{cm} + F r \theta$. The first term is the translational energy increase and the second is the ...


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What is the proof of this? The proof of Chasles’s Theorem relating to the rotational and translational displacements of a rigid body is done in very many texts including Appendix 20A of this document produced by MIT. In essence this shows that, if a force whose line of action does not pass through the centre of mass of a rigid body, the applied force is ...


4

You can apply a force of the same magnitude in both cases, and the work will be different. The reason is that the displacements are different. The CM will displace the same amount, so the final kinetic energy of the CM will be the same in both cases. But in the case of the extended object the point of application displaces more than the CM, that is why the ...


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Angular momentum itself does not contribute to the invariant mass, however an object with angular momentum will have KE in its center of momentum frame, and this KE is part of the invariant mass. It is the KE that provides this “additional” mass, not the angular momentum. So, for example, two objects with the same angular momentum but different amounts of ...


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@MichaelSeifert has a very nice answer. I just want to describe it from a different angle here. A Calculation If you only want to investigate if contact is lost at some angle $\theta$, then in this problem it can also be done in the following way: Only the lower end of the rod is in contact with the ground. So for the rod to remain leave contact with the ...


1

Although the static friction does no work it does apply a force to the bottom of the rod (which accelerates the center of mass horizontally), and a torque. The normal force also produces a torque relative to te center of mass. Both of these torques change as the rod falls, and cause changes in the angular momentum. (The angular acceleration is a variable, ...


1

The stool can rotate freely only about the vertical axis. It can and does provide counteracting torque to any component of L perpendicular to that. That allows it to provide the horizontal components of torque necessary to cancel the L2 in the diagram. Physically, the stool is what’s preventing the person from rotating in pitch (head forward, feet back)...


4

to see what happens you have to write the equation of motions and then simulate the equations. we have two generalized coordinate $x$ is the translation on the floor and the rotation of the rod. starting with the position vector to the center of mass you get: $$\vec{R}=\left[ \begin {array}{c} l\sin \left( \varphi \right) +x \\ l\cos \left( \varphi \...


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The technique to use in problems like this is to assume that the rod remains in contact with the table, and to then try to figure out whether the normal force ever switches sign for some angle $\theta$ as the rod falls. If it does, then the rod's lower tip will have to leave the table, as a "frictionless table" cannot pull the rod downward; it can only ...


1

You can start building your intuition from special cases (like you have done for $v_A = v_B$). The next special case to consider is that only one of the side pulleys is giving rope, i.e. let's say $v_B = 0$. Then it is easy to see that the main pulley should be descending at $0.5v_A$. How about the rotation? The main pulley and the pulley A have to pass the ...


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This is a nice question. I agree that the standard textbook reflection on this problem generally leaves you wanting more (but this is perhaps unavoidable). The basic question one grapples with at the start is $-$ how come suddenly we've to switch to torques and angular momentum when it comes to rigid bodies, as opposed to mere forces in case of point ...


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Because each point in a rotating object is going in a different direction, the vector representing angular velocity was chosen to be along the axis of rotation, the only direction which characterizes the system as a whole. Torque as a vector is defined along the axis for the same reason and because we want the torque vector to be in the same direction as ...


2

Torque is defined as $\quad\vec{\tau}=\frac{d\vec{J}}{dt}$ where $\vec{J}$ is the angular momentum of the object. The angular momentum is defined as $\vec{J}=\vec{r}\times \vec{P}$. Then $$ \vec{\tau}=\frac{d\vec{J}}{dt}=\frac{d(\vec{r}\times \vec{P})}{dt}=\frac{d\vec{r}}{dt}\times\vec{P}+\vec{r}\times\frac{d\vec{P}}{dt} $$ but $$ \frac{d\vec{r}}{dt}\...


-1

The answer is because we can slide a force vector along the line of action and it does not change the system. Hence the location of a force along the line of action is not important and there only thing that matters is the moment arm of force. Consider a force vector $\boldsymbol{F}$ acting through a point located by the position vector $\boldsymbol{r}$. ...


15

It doesn't have to be thought of as cross product. It's just very convenient to think of it that way, so we teach it first. Indeed, even when I apply it in my job, I think of it as a cross product. But first, your question about why the lever arm appears in the equations. Informally, we need to account for the length because a longer lever arm gives you ...


1

How about imagining a painted marker on each cable which is just above the pulley. The marker for each cable will move with the speed of the respective cable. The only thing that matters as far as the main pulley rotation is the difference in height between those two markers etc. I hope this helps.


2

(a) "why do we consider the length between the axis/point of rotation while calculating torque?" We can calculate torque about any point, O, that we choose; it doesn't have to be a physical axis of rotation. But it's often more useful to calculate torque about a possible physical rotation axis, for example when thinking about what torque we need to apply on ...


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Perhaps the best way to explain torque is with respect to the use of a common tool: The wrench. If we want to tighten or loosen a bolt we may use a wrench. The first thing we realize is if the bolt is frozen in place, we are more likely to free it if we use a wrench with a longer arm. This is because the force we apply to the wrench arm produces more torque,...


0

Several people have suggested that the contact point will act as an axis. I don't see how this is the case, since it's in no way constrained. For example, the axle of wheel is constrained by being attached to a car or bike, thus if you push on the rim that force gets turned round the axle. If I push a person (standing on normal ground) near his CoG, that ...


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Roll a toy car along a surface, at any angle, and the wheels rotate. That's because friction is applying a force at the contact point between the wheels and the surface. Now move it parallel to the surface, a few cm away. The wheels do not rotate because there is no force at the contact point. The reason there's no force is because there's no friction (...


1

One way to see that it couldn't depend directly on $r$ is that it would imply an object with radial symmetry has zero moment of inertia about an axis through its center of mass. Any torque applied would then impart an infinite acceleration.


0

The motion of the center of mass and the rotation about the center of mass are independent of each other. In fact, this is what makes the center of mass of an object special. You define momentum and angular momentum about the center of mass and they are completely independent of each other. $$ \begin{aligned} \boldsymbol{p} & = m\, \boldsymbol{v}_{\rm ...


1

Welcome to Physics Stack Exchange! Rotation and translation are separate processes. The energy of a translating process is $$ KE = \frac{1}{2} mv^2$$ where KE = kinetic energy and m,v are the mass and velocity of the object. The $$E_{rotational} = \frac{1}{2} I\omega^2$$ where I = moment of inertia about axis of rotation and $\omega$ = angular velocity. If ...


1

$1^{st}$ method The torque is only generated by the perpendicular component of force acting along the radius vector (as can be seen from this formula $ \mathbf \tau = \mathbf r \times \mathbf F $) $2^{nd}$ method You may know that $v \neq \omega r$ but $\mathbf v = \mathbf \omega \times \mathbf r $. Therefore $\mathbf a = \mathbf \alpha \times \mathbf r $...


2

When you pull your arms in you aren't pulling them directly towards the centre, because you're rotating as you're pulling them in. This is where the force comes from that actually makes you spin faster. You should definitely watch this video where he explains exactly this. Skip to 10m in if you're in a hurry but the whole video is well worth watching. I ...


1

Can we explain the above case with the help of just forces without using any result like "Angular Momentum Conservation? I think that will give us more insights into what is exactly happening. Well, yes and no. Because with angular conservation laws, you can just say that $L = L$, which means that $I_1 \omega_1 = I_2 \omega_2$, and then you can do some ...


2

so we can conclude that the man has angular acceleration without any external torque, which is an apparent contradiction of the terms, so how do we reconcile the case with the concept? We reconcile it with the law of conservation of angular momentum. The angular velocity of the skater increases when drawing in the arms in order to conserve angular ...


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The definition of torque is not $\tau=Id\omega/dt$. We can't even define things like $I$ and $\omega$ for rotation that isn't rigid. The definition of torque is $\tau=dL/dt$. So yes, it is possible to have an angular acceleration without an external torque. Your example shows correctly that this can happen.


1

Even if $\vec{F_1}+\vec{F_2}+\dots=0$, $\vec{r_1}\times\vec{F_1}+\vec{r_2}\times\vec{F_2}+\dots$ may not. One example, consider such lever system : Weights cancels each other : $$ W_U + W_D = 0, $$ however, net toque is $$ \tau = WL_1 - W(L_1 + L_2) = -WL_2. $$ So it is not zero and directed downwards, given that lever will not break at point A


1

The ball will slide. You mistake was to choose an 'accelerating axis' (The point of contact through which the axis passes is accelerating). Note that the you can only form torque equation about the axis which are stationary or translating with constant velocity. The beauty of centre of mass is that torque equation can be applied to an axis passing through ...


1

But that won't always work. Even if $\vec{F_1}+\vec{F_2}+\dots=0$, $\vec{r_1}\times\vec{F_1}+\vec{r_2}\times\vec{F_2}+\dots$ may not. Clearly this is true. If the sum of the forces being zero always meant the sum of the moments (torques) is also zero, you would be able, for example, to determine static equilibrium requirements for beams based only on the ...


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The ball will roll. The centre of mass of the ball is not vertically above the point of contact, so there is nothing to prevent it from falling vertically. The two conclusions you mention are not contradictory- you have simply misunderstood their implications. There is no turning moment about the centre of the ball. There is a turning moment about the ...


1

I think much of the confusion comes from the notion that a moment (or torque, I use these words synonymously) might have an axis or a place, which is false. In classical mechanics, such a torque is associated with any rigid body, rather than with a specific place. Strict method Usually, you would cut the rigid body free from its surroundings, thus ...


1

This recent question and its answers are relevant. Setting the torque about the point of contact with the ground to zero means that you are assuming the angular momentum, $L=m\boldsymbol{r}\times\boldsymbol{v}$, about this point to be constant. In other words, the linear velocity of the yo-yo is assumed to be constant. In your second case, setting the ...


4

to see what happens, lets look at the equations of motion: $$m\,\ddot{s}+F_c-m\,g\sin(\alpha)=0\tag 1$$ $$I_b\,\ddot{\varphi}-F_c\,R=0\tag 2$$ case I: Ball is rolling without slipping: $$\ddot{s}=R\ddot{\varphi}\tag 3$$ you have three equations for three unknowns $\ddot{s}\,,\ddot{\varphi}\,,F_c$ you obtain: $$\ddot{\varphi}=\frac{m\,g\,\sin(\alpha)\,...


45

...the torque exerted by $N$ is zero but the torque exerted by $mg$ is non-zero. This means the ball must roll... Actually, it means that the angular momentum about that axis must increase. That is not the same as rolling. If the axis is through the center of mass of the object then the only way for the angular momentum to increase is through rolling. ...


20

So, what exactly will happen to a ball kept on a frictionless inclined plane - will it slide or roll? Frictionless means the surface of the incline cannot exert any torque on the ball. By Newton's second law, that means the state of rotation of the ball remains unaltered, specifically: if the ball was spinning at angular velocity $\omega$ then it will ...


0

Assuming the sticks are in the shown position (but touching) when the lower stick fuses with the upper one, then the top end of the lower stick has been coming toward you, and imparts some of that motion to the center of the upper stick. But, the right end of the upper stick has been moving away from you and will continue to do so. The point where these ...


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