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4

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.


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 ...


3

At the instantaneous moment shown in the diagram, we can write: $$2R\alpha_{ring}=a_{disc}$$ as both are in pure rolling. This also tells us that the point on the ring where the thread is attached has an acceleration $=2R\alpha_{ring}=2a_{ring}$ so we find that: $$a_{disc}=2a_{ring}$$ Note that when the string moves to another position this will not be true, ...


3

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.


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.


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.


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 ...


2

In your scenario, angular momentum $m v r$ is preserved (because your pulling force is radial, with no tangential component). So if you reduce $r$ by half, $v$ must double, and since $\omega = v/r$, it increases by a factor of four. Note this means in a small amount of time that the area swept out by the string is proportional to $v$ and $r$. Since they ...


2

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) ...


1

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} \\ ...


1

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 ...


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 ...


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 ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


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 ...


1

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 ...


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.


1

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

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 ...


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.


1

As the ring moves forward, the string unwinds from it. When the ring has completed one revolution, every point on it has moved forward by the distance of its circumference. The string has unwound by an amount equal to the ring's circumference. So while the centre of the ring has moved forward by one ring circumference, the end of the string (where the ...


1

Okay, so I figured it out myself. Here's what I think: Take $P$ to be the point on top of the ring, where the string is attached. Now, two things contribute to $P$'s acceleration : the acceleration of the centre of mass of the ring, and the acceleration due to the angular motion. So, $a_P = a_{ring} + \alpha \times R$, where $\alpha$ is the angular ...



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