New answers tagged

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When you are considering torque on a part of the wheel (I am assuming you are considering the torque about an axis passing through the center of the wheel), you also have to take into account the torque due to weight of the section of the wheel which is no longer zero as you had in the case of the whole wheel (assuming that mass distribution is uniform). So ...


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The angular momentum vector is: $$\vec{L}=I\,\vec{\omega}$$ if you take the components of the angular velocity $\vec{\omega}$ in B-frame and the Inertial tensor $I$ also in B-Frame you get the components of the angular momentum in I-frame with this equation: $$\vec{L}_I=R\,I_B\,\vec{\omega}_B\tag 1$$ where R is the transformation matrix between B-frame ...


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Given the current form of the equation, I have to make a few assumptions about the currently-missing context: $\mathbf{L}$ is the angular momentum in a fixed frame. $\mathbf{M}$ refers to the rate of change of angular momentum in a frame that is rotating with angular velocity vector $\vec{\omega}$. The $\vec{\omega}$ in the final term of the equation is ...


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It is true that at any instant the acceleration of a vehicle is directly due to the torque supplied by the engine. And in a vehicle with direct drive (one gear) there is a 1:1 relationship between acceleration and torque $$ a=\frac{T_{\rm wheel}}{m\,R_{\rm wheel}} =\frac{T_{\rm motor}}{m\,R_{\rm wheel}} \tag{1}$$ Electric Motor Torque produced isn't ...


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The rate at which a massive object can be accelerated depends on its mass and on the power available to accelerate it. In the mechanical realm, power is the product of an effort variable (torque) and a flow variable (RPM). In this context, the objective of a transmission is to impedance-match the torque/RPM characteristic of the prime mover (the engine) ...


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Slip is not "impending in the direction in which the wheel rolls". Slip is impending in the opposite direction to which which the present forces are pushing it. And the force to consider here is gravity; it is trying to make the contact point slip by pulling the ball downwards, so static friction must pull upwards to prevent the contact point from slipping. ...


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First, think about how the surfaces would slip without friction. In this case the wheel would slide down the incline without rolling. Static friction will therefore try to prevent this, and so must point up the incline. Another way to think of it: you have assumed rolling without slipping. The only force that exerts a torque about the center of mass of the ...


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The moments tell you where a force is applied. The force vector will give you the magnitude and direction of a force, but not where the line of action of the force is. This is why measuring forces at different points always produces the same result, but the resulting moments change with location. To get the line of action you need the moments. If at point ...


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On falling mass there is a force mg acting downwards and a tension force acting upwards to calculate tension Write the expression of torque due to tension on pulley and equate with I*alpha Considering pure rolling acc will be product of radius and angular acceleration The last equation can be made as mg-T =ma where m is mass t is tension


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Well, considering vertical motion with acceleration $a_y$ (in +ve Y-direction) then $$\sum F_y=ma_y\iff N_f-F_g=-ma_y<0\iff F_g>N_f$$ Similarly, for horizontal motion towards right, $$\sum F_x=ma_x\iff N_w-F_f=ma_x>0\iff N_w>F_f$$ Considering rotation of ladder anticlockwise with angular acceleration $\alpha$ about C.M. in plane of paper $$\...


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This answer isn't yet about gyroscopes. What this answer does do: I support that the concept of vector angular momentum doesn't help with understanding gyroscopes. About how angular momentum is commonly represented: Our space has three dimensions of space. In a space with three spatial dimensions each plane has a single line that is perpendicular to that ...


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At least for me, gyroscopic precession never made intuitive sense until I watched this video; https://www.youtube.com/watch?v=n5bKzBZ7XuM&t


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Internal forces are forces which do not cause any change in the acceleration of center of mass. That doesn't mean they can't accelerate an individual mass with respect to the center of mass. Thus, they can be perpendicular to the rod.


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For any couple, the axis of rotation is always through the COM and is truly parallel to the torque. Torque is a vector whose direction is determined by the right hand rule. Depending on the orientation, the torque can be pointing either up/down or into/out of the reference plane. And this direction will always be parallel to the rotation axis. To understand ...


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The torque produced by a selected force relative to a selected point is only zero if the line of action of the force passes through the point.


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In mechanics, a couple refers to two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action. A better term is force couple or pure moment. Its effect is to create rotation without translation or, more generally, without any acceleration of the centre of mass. Your one statement is correct that the axis of rotation ...


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You are actually asking two questions, in the first you have constant torque from a motor, and in the second, when you use a crossbar, you have constant force, which is what we humans can do. In the second case the torque you can do increases with the length of the bar so you can unscrew the screw.


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You have to use gears to increase torque. If the shaft's input end is rotated by $\Delta\theta$, the output end also rotates by the same amount $\Delta\theta$. If the input torque $\tau$ did a work $\tau\Delta\theta$, conserving energy gives $\tau_{out}\Delta\theta = \tau\Delta\theta$ Thus $\tau_{out}=\tau$ By changing the shape you have actually ...


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I'm assuming that you're turning it around the square bit? That increases the distance from the pivot which I'm assuming is the middle of the circular arrow so it should.


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Any combustion engine is capable of operation at a variety of RPM's and torque output combinations that depend on the nature of the load that the engine is driving. Since delivered power is the product of torque and RPM, this means that there are many different combinations of the two across the engine's operating regime that will yield the same net power ...


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Both the equations are completely consistent with each other. Equating both the equations (assuming that $F$ is the only force acting on the body, thus we can use $F=ma$) $$Fr=I\alpha\Longrightarrow mar=kmr^2 \alpha\Longrightarrow \boxed{a=kr\alpha}$$ This equation is true for any general body as long as there's only one force acting at a distance $r$ from ...


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Your diagram is pretty much clear, wall normal force at contact point A is : $$ \vec{N}_A = - (\vec{F}_1+\vec{F}_2+\vec{T})_{\perp\,\text{wall}} $$


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"Performance" is a pretty broad term. For example, it can relate to how fast the bolt (the arrow) is launched, or the repeatability of the bolt's trajectory. Plastic bolts are not a very good idea: most plastic can deform permanently, which will lead to a wildly variable trajectory. You might do well to buy some fiberglass or graphite fiber fishing pole ...


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A better crossbow design than a fixed wooden rod and an elastic band is a tough piece of string connecting to "sprung" arms that are pivoted at the join. They can be sprung by winding up elastic as the string is pulled back. This makes for a better crossbow as it is much more effective at storing elastic potential energy as the mass of the arms move as well ...


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I guess there is a slight mistake in the diagram. If the bicycle is not turning, then the friction will because there is no tendancy for relative motion in the horizontal direction (towards the left, that is). So all that exists is a couple-the Normal force and weight with two different lines of action, hence causing the rider to fall. In case a force due ...


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There are actually several questions wrapped up in what you have asked. If you don't mind, I'd like to sort them out, and tidy them up a bit: Why is it that leaning into a turn while on a bicycle results in the bicycle changing directions rather than falling over. Put another way, why doesn't a bicycle fall over when you lean into a turn while riding. ...


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Angular momentum of a system is only conserved about an axis if all the external forces are passing through the said axis (because then no forces produce any torque about the said axis). In your example, you're considering the bullet + cylinder as your system. The forces in the scenario are the forces between the bullet and the cylinder and the friction ...


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Torque is best defined as the work per unit angle of rotation which can be done by a force acting in a manner which tends to cause a rotation. With a longer lever arm, your applied force moves through a greater distance and can do more work on the nut or bolt for the same amount of rotation.


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I hope you are somewhat familiar with linear algebra and rotation matrices. To find the mass moment of inertia 3x3 matrix along the world coordinates, from known 3 principal values along the body frame you do the following: $$ \mathbf{I} = \mathbf{R} \, \mathbf{I}_{\rm body} \, \mathbf{R}^\top\tag{1} $$ where $\mathbf{I}_{\rm body} = \pmatrix{ I_1 & &...


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As I see it, your question has four parts. 1. Moments of inertia about $\hat i$, $\hat j$, and $\hat k$ The moment of inertia of any body can only be defined along an axis. It's defined as: $$\int r^2 dm \tag{1}\label{1}$$ where $r$ is the distance of $dm$ from the axis. This is usually different for different axes, except for cases where some axes are ...


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Hint: There are two equations for static equilibrium, (1) the sum of the forces equals zero and (2) the sum of the moments (torques) equals zero. For the first equation you only need the persons weight. For the second equation you need the location of the person's center of mass. You will need to apply the second equation twice, once for each diagram, ...


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You have two choices if you want to consider a point that is not the CoM. consider the point fixed in an inertial frame. Under this method, you are doing an instantaneous calculation. The object will move away from that point, but the point itself is not accelerating. The main problem is that if the CoM of the object is not in line with the point, its ...


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You already have your answer: But, it takes energy to curl up and un-curl. Your muscles do work to move your body. That is where the "chemical energy" you are asking about goes.


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Angular momentum is given as $\textbf{L}=\textbf{r}\times\textbf{p}$. Thus, we obtain that torque, defined as $\frac{d\textbf{L}}{dt}$ is $\textbf{r}\times\frac{d\textbf{p}}{dt}=\textbf{r}\times\textbf{F}$. As you know, the $\frac{d\textbf{r}}{dt}\times \textbf{p}$ factor vanishes because it is the cross product of two parallel vectors. Now, the reason as ...


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The formula for torque is ultimately set by an arbitrary choice in the way that we define angular velocity. We have chosen to define angular velocity according to the right-hand rule - in other words, we have arbitrarily chosen that counterclockwise motion corresponds to an angular velocity vector pointing upward. Defining angular velocity this way means ...


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The yarn can be modeled, very roughly, as a torsion spring. The torque $\tau$ required for a particular angular deflection ("twist") $\theta$ is given by: $$\tau = \kappa\theta$$ The constant $\kappa$ is called the torsion constant. For an object with a uniform circular cross section, $\kappa$ is related to three quantities: the length $L$ of the yarn, the ...


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What I'm asking is if we can apply the brakes on the car to stop the wheel but without any static friction appearing. No. Or does static friction appear as soon as we have any counter torque acting on the wheel? Yes. Per Newton's third law, a static friction force will be exerted by the road on the tires as soon as there is a counter torque ...


1

Imagine an incline with constant inclination. A sphere or cylinder is rolling down it due to gravity: Imagine also that the top half of the incline provides enough friction so that the object rolls without slipping. And the bottom half of the incline is completely friction-less. Rolling without slipping means simply: $$v=\omega R$$ where $v$ is velocity,...


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A sphere may be rotating (in any direction) as it moves across a friction-less surface, but the term rolling generally implies an instantaneous static contact between the surface and the bottom of the sphere.


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