New answers tagged

1

does it also leads to ... No, it doesn't. A simple Counterexample: Consider the figure below (the bar $\textrm {AB}$ is on a plane parallel to $\textrm {xy}$ plane) We have $\Sigma \vec F=\vec 0$, but, if we calculate vector sum of torques about point $\textrm A$ we will obtain $\Sigma \vec M_A=F (\overline{AB})\vec k\neq \vec 0$ ($\vec k$ is the ...


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The friction force doesn’t do work because the contact point of the disc with the ground changes. It is not same for a distance. At each moment, there is a new contact point. Hence, for all friction forces (for all contact points), $d=0$ and $W=Fd=0$.


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Imagine that friction between wheels and road disappeared in one second. Then the car would just go away moving straight along tangential direction, right? In the life it doesn't, consequently there is centripetal acceleration that keeps the car on the circle way. Consequently there is some force doing that. It is friction.


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The force of friction acts both towards the centre of the circle and opposite the velocity vector of the car. Strictly speaking, the diagram you have does not show all forces acting on the car but it is enough for purposes of explaining the circular motion. As the text also explains, circular motion always requires a force pointed radially inwards because ...


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If you see on the cylinder as a wheel then note that its center moves twice slower than top point attached to the block. Same is for acceleration. In each moment the cylinder rotates around the point of its touch to the table, so radius from touch-point to the center is twice less than radius to the cylinder top point.


2

I think you are asking about how to transform various vector quantities between points attached to a rigid body. Here are a rundown of the rules of transformation between an arbitrary point A (located at $r_A$) and the center of mass C (located at $r_C$). $$\begin{align} v_C & = v_A + \omega \times (r_C - r_A) & & \text{linear velocity at C} \\ ...


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When the wheel comes in contact with the belt friction will act as there will be relative motion between the belt and the point of contact. Now friction will tend to act on the belt opposite to the velocity of the belt until slipping ceases. To find the work done by the external agency lets consider the energy changes : 1) The K.E of the wheel increases. 2) ...


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In your example you have the total load applied as $$ F = \int \limits_0^\ell w\, {\rm d} x = w \ell $$ where $w$ is the linear force density (in Newtons per meter). To get the torque you just include the position of the force $$ \tau = \int \limits_0^\ell x\, w\,{\rm d} x = \frac{1}{2} w \ell^2 $$ The rule for distributed loading is to find the total ...


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In the case of the gate, F=200N is the distributed force, which is applied from x=a=0m to x=b=2m from the axis. The force applied on an element of length dx is (F/L)dx Newtons where L=b-a. The torque on the gate due to the force on this element is dT = (F/L)xdx. We integrate this from x=a=0m to x=b=2m to get the total torque : $T = [\frac{F}{L}\frac12x^2] ...


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A force applied uniformly to a uniformly dense object (or with a uniform-force-per-unit-mass even if the object is not uniformly dense) is equivalent to the same total force applied at the center of mass of the object. The equivalence means the same total force and the same total torque. In particular, if the center of mass is chosen as the origin of the ...


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There is no need to calculate the moment of inertia I, which could be quite difficult for an irregular lamina. You can find both I and g from the same experiment. The period T of small amplitude oscillations of a compound pendulum is give by (see Source below) : $(\frac{T}{2\pi})^2 = \frac{k^2+h^2}{gh}$ where k is the radius of gyration (related to ...


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Of course it can. Angular momentum (the rotational analog to linear momentum) can be expressed as an axial vector (sometimes called pseudovector), which means a quantity that is similar to a regular vector, except for the fact that it changes sign under improper rotations (such as reflections). I suppose you're refering by "wobble" to nutation, i. e., a ...


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Well, this will certainly depend of the shape and mass distribution of your pendulum. For instance, for a pendulum made of a rod and a thin cylindrical disk, the moment of inertia of the rod (about its own symmetry axis) would be $\frac{1}{12}M_rL^2$, and the moment of inertia of the disk $\frac{1}{2}M_sR^2$. If you have a solid sphere instead of the disk, ...


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Take an arbitrary cross section with normal stress $\sigma(x,y)$ as a function of location and sum up all the forces and moments $$\begin{align} \sum F & = \int \sigma \,{\rm d}A \\ \sum M & = \int y \sigma\,{\rm d}A \end{align} $$ You can find the neutral axis when only bending stress exists (no axial forces) by taking the first equation $\sum F ...


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In beam bending, the bending stress in the cross section is proportional to the location where the stress is calculated from neutral axis of the beam, where by definition the bending stress = 0. The neutral axis of the cross section is found by calculating the first moment of area of the cross section, and then finding the centroid using that moment. In ...


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The PE of the particle is converted into the KE of the particle $\frac12mv^2$ plus the KE of the disc $\frac12I\omega^2$ which is also moving.


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The $\frac m 2 v^2$ term is the kinetic energy of the "small particle". The $\frac {I}{2}\omega^2$ is rotational kinetic energy of the disc of mass $M$. You are just counting the kinetic energy of each mass once.


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I assuming that you wish to find the distance from the thrower as a function of time. Suppose the thrower is at the origin, and the boomerang moves anti-clockwise in a circle of radius a with centre at (0,a) with angular speed $\omega$. Then the position of the boomerang at time t is (x,y) where $x = asin(\omega t)$ $y = a(1-cos(\omega t))$. The distance ...


1

Let $x_1,\theta_1$ etc. with subscript 1 be coordinates of the upper mass, and variables with subscript 2 be coordinates of the second mass. So,the Lagrangian can be constructed as: $$ L = 1/2M\dot x_1^2 + 1/2 \ I\dot \theta_1^2+ 1/2m\dot x_2 + 1/2m\dot y_2^2 - mgy_2$$ Now, $$ y_2 = L \ cos(\theta_2) \\ x_2 = x_1 + L \ sin(\theta_2)$$ I think that accounts ...


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Consider the arm to be massless and of length $L$. Two torques now act on the wheel: $F_1L=mgL \sin \theta $ And: $F_fR=\mu NR$ With: $N=Mg+mg=(M+m)g$ So: $F_f=\mu (M+m)gR$ Newton's second law (here on rotation) tells us: $$I\alpha=-mgL\sin \theta+\mu (M+m)gR$$ Or: $$I\frac{d^2\theta}{dt^2}+mgL\sin \theta=\mu (M+m)gR$$ If we accept the small ...


1

This is one of those three part dynamics questions. For the first part you need to use energy conservation to work out the horizontal speed of the person just before hitting the pole. The second part is the application of the conservation of angular momentum about the pole's pivot point when the person grabs hole of the pole. Note that the collision between ...


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In a vacuum the rod will keep its orientation because of the equivalence principle, but with air resistance the heavier part will tilt toward the ground because its mass and therefore force is higher. The force depends not only on the mass of the planet but also on the mass of the falling object, and since air resistance is also a force that acts on the ...


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A ball of radius $d$ rolls (w/out slipping) on two surfaces. One surface moves with $v_1$ and the other with $v_2$ speed (in the same direction). The linear velocity of the ball center is the average velocity $$v_{ball} = \frac{v_1+v_2}{2}$$ and the angular velocity proportional to the speed difference $$\omega = \frac{v_2-v_1}{d}$$ These quantities are ...


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The gravitational force of the first mass is $$F_1 = m_1\cdot g/ r\cdot sin(\varphi)$$ and on the second mass $$F_2 = m_2\cdot g/ r\cdot sin(\varphi+\pi)$$ which is in total $$F_g=F_1+F_2=m_1\cdot g/ r\cdot sin(\varphi)-m_2\cdot g/ r\cdot sin(\varphi)$$ which cancels to $0$ if $m_1=m_2$. So if your disk is aligned vertically and the two masses are ...


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If the water bucket is not accelerating there should not be any resultant force on it. The bucket is moving at constant speed, so all forces acting on it cancel: it has gravity acting with $mg$ on it and tension equal and opposite to gravity. But, F in the equation for power is not the net force, it is the driving force: $$P = F_{driving} v$$, i.e. the force ...


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The only way that friction can appear is for there to be tension, since it is tension that will give rise to the normal force needed for friction. Now if we note that the friction must result in a difference in tension between the left and right strings (if the masses are different) then there will be a continuous change in tension. The normal force at ...


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The rocket nozzle is mounted on a bearing that can tilt a small amount in two directions - allowing it to steer the direction of the exhaust and so the direction of thrust. Thrust vector nozzle The gimbal linked in the article is for the sensor platform used to measure the direction, although now you would use a solid state 3d acceleration sensor instead of ...


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The meaning you quote is only one of several. From the same dictionary, others which are now obsolete or not often used are 3: importance in influence or effect 4 obsolete : a cause or motive of action and it is from these that the scientific meanings derive : 6a : tendency or measure of tendency to produce motion especially about a ...


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If the pulley is free to rotate and perfectly frictionless then the tension in the string is the same throughout, from A to D. So the tension in BC is not zero. You are correct in your comment stating that the net force on BC is zero, because the tensions in AB pulling right and in CD pulling left are equal. However, the net force acting on section BC is ...


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In the language of physics, a moment is a physical quantity which accounts for how a physical property is located or arranged. We do NOT use the colloquial meaning of the term moment. This "moment" we are referring to is derived from the latin word momentum which is also a physical quantity equal to the product of the mass and velocity.


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The moment of a vector $\vec v$ applied in the point $\vec p$ with respect to the pole $\vec q$ is by definition $$\vec M = (\vec p - \vec q) \times \vec v$$ With $\times$ vector product.


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When a top rotates, it rotates about its centre of mass. The centre of the mass is a point on the axis of rotation. Since the axis is also stationary as is the centre of mass, therefore all the points in the axis are eligible to be considered fixed about which the top is rotating. Besides,I would prefer to use the term axis instead of a fixed point.


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Your first calculation is correct. Nothing wrong with it. The mystery for me is why you go to so much trouble trying to disprove it!


1

You have made some errors in you calculation of distances. Let $$ \mathbf{r}^{\prime}_{k}=\mathbf{r}_{k}-\mathbf{r}_{_{CM}} $$ Then \begin{aligned} r^{\prime\;2}_1 &= \Big(\frac{24}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 = \frac{1872}{121}\quad \text{(upper left particle)}\\ r^{\prime\;2}_2 &= \Big(\frac{20}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 ...


2

shouldn't we use the parallel axis theorem ... to compute the moment of inertia? You could...if you already had the moment of inertia of the object about its center of mass. Since you don't, it's far easier to simply sum the moments of inertia about the $z$ axis.


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It may be 5 years on from the original question, but it's a shame to have only a single accepted answer, which is just plain wrong. Though it's true, that the human body has a "natural intuition for physics", this is only within a pretty wide margin of error (which gets narrower with practice in spear-throwing). It is not a flick of the wrist, which makes ...


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First, there are a few different methods developed to solve this kind of problem, but it's highly dependent on your background knowledge (calculus), and experience with these kinds of problems. This approach might be over-detailed for some mechanics problems, but this approach is fairly general, so should usually work. Identify any clues in the problem as ...


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You are almost correct. The linear acceleration of the end of the rod is $\alpha L$ perpendicular to the rod. The component of this acceleration in the direction of the string is $\alpha L\sin A$ where $A$ is the angle which the string makes with the rod (not the angle which the rod makes with the horizontal). So at the instant shown in the diagram the ...


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A pure moment cannot be created by one force. In real life, producing pure torques is almost impossible. There will always be a non-zero net force applied also. In your examples the torque about the center of mass is $\tau_C = 2 r F$ in both cases, but on the second case you also have a net force applied $2 F$ that changes linear momentum also. So your ...


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The error lies in the theoretical confusion between forces/torques and energy. The kinetic energy is linked to the motion generated by forces and torques which are the causes of the motion itself. Understanding the energy value in the two situations results impossible without knowing the time course of the applied force, being the energy, and so the work, ...


1

If there is no friction the energy you put in initially will be conserved and the flywheel will rotate forever, then you don't need to put in any extra power to keep it running.


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The rod will only rotate upwards due to the force that is normal to it's length, acting from the tether on the rod, $$F_{t,\perp} = F_t\cos\theta$$ Here $F_t = (2m)g$ is the force from the tether, equal to the weight of the block, and $\theta$ is the angle. The rod also feels a downward force, gravity, that has a normal component $$F_{G,\perp} = ...


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Kinetic energy is $$K = \frac{1}{2} \vec{\omega} \cdot [I] \vec{\omega} $$ when the 3×3 mass moment of inertia matrix $[I]$ is expressed in world coordinates. Remember $$[I] = [R] [I_{body}] [R]^\top$$ is how body inertias is transformed into world inertias. You seem to apply a scalar mass moment of inertia to a vector rotation. If you are careful with the ...


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For First Case: the rotation produced by the torque at the centre of wheel will rotate the wheel in the clockwise direction, but here friction is present, as friction opposes the motion of particle that's why it acts in anticlockwise direction and helps the body to move. For Second Case: here $mg\sin\theta$ will act along the line of centre of mass, which ...


2

In general, the change in angular momentum resulting from a change in moment of inertia depends on how the change is implemented, and to some extent your perspective. In physics, you can think of global conservation laws as constraints that feed into your interpretation of a system. Consider the simple problem of determining the change in linear momentum ...


4

The principle of conservation of angular momentum says that angular momentum remains conserved unless an external torque acts on it. The net torque on a body is defined as: $$\vec{\tau\,}=\dfrac{\mathrm d\vec{L\,}}{\mathrm dt}$$ We can clearly see from this definition that since external torque on the body is zero, the angular momentum is going to remain ...


2

Remember that the variation of the angular momentum equals the external torque. If there are no external torque (as in your case), the angular momentum is conserved.


0

You have my sympathy; precession is little understood. Try it this way! i) remove the front wheel from a bicycle and mark a spot on the tyre. ii) hold it by the ends of the axle between your hands with the mark at "top dead centre". iii) set the wheel spinning as though the machne was progressing forward (very slowly) Note that the spot travels ...


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Yes. Precession is the answer but it seems to be a sbject which is poorly understood and insufficiently researced. It becomes particularly interesting when two gyroscopes are mechanically-coupled: e.g. a bicycle - and closed feeedback-loops come into play.


1

I would suggest that the major reason for the ball becoming stationary is its inevitable interaction with the air - i.e. friction, resistance to being parted and eddy currents.



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