New answers tagged

2

Fix a reference frame $R$ with axes $x,y,z$ and origin $O$ and suppose that the solid body $B$ is moving with a planar motion, let's say, parallel to the plane $x,y$. If $O'$ is a fixed point of $B$, the velocity of a point $P \in B$ in $R$ satisfies $$\vec{v}_P(t) = \vec{v}_{O'}(t) + \vec{\omega}(t)\times \vec{O'P}(t)\:.$$ Since the motion is planar $\vec{\...


0

For each object for energy conservation you will have $\Delta\text{KE}=-\Delta\text{PE}$. Since both of these terms are directly proportional to the mass, the mass variables on each side of the equation will cancel out. Therefore, you do not need to know the mass of each object to solve this problem. Also, in this case speed means the translational speed of ...


5

First, note that you can't "misdefine" something. Angular acceleration is defined to be the time derivative of angular velocity; that is it. Instead of questioning the definition then, you should be questioning your understanding of the relationship between this definition and other physically relevant definitions. The error is in assuming that the moment ...


1

The following analysis is performed in an inertial "lab/ground" reference frame. Let our system be a collection of point particles labelled by index $i$: $\{m_i,\mathbf{x}_i,\mathbf{v}_i,\mathbf{a}_i\}$ and let $P$ be a general point that may or may not be in motion with respect to the inertial frame. The angular momentum with respect to point $P$ is ...


0

yes the situation is indeed a non inertial one, but if we consider a inertial frame such as with respect to(wrt) inclined plane, we don't get any axis to apply torque (using ICOR) it comes out that point of center of mass(COM) is the axis of rotation, so we apply torque wrt COM frame which is indeed non inertial, torque is produced by friction in this case, ...


0

I think the reason velocity is root3 m/s and not 1m/s is that when both body are doing pure translational or rotational motion then V 1/2 = -V 2/1. But in this case the second body is not actually doing a rotation motion but in its frame of reference a component of angular velocity is there just like in the case when 1 and 2 both are performing rotation ...


0

The solution for this kind of differential equation $${\frac {d^{2}}{d{t}^{2}}}q(t) +{\omega}^{2} q(t) =0$$ is: $$q(t)=A\,\sin(\omega\,t+\varphi)$$ where: $\omega$ eigen angular velocity $\varphi$ phase $A$ Amplitude the frequency of the sine wave is $f=\frac{\omega}{2\pi}\quad [\text{hz}]$ and the period of the wave is $T=\frac{2\pi}{\omega}\quad [\...


2

...you need $\omega = \sqrt{\frac{k}{m}}$ to even show (which the book doesn't) that $x(t) = A\sin(\omega t)$ is a solution. Actually, you don't. You can go through the whole process of deriving the differential equation that governs a simple (or physical) pendulum, approximating it, and solving it without ever using the symbol $\omega$ or talking about a ...


0

For a wheel, I would use rotational variables, starting with angular velocity, w = 2πf (in radians/sec). Then wf = wo + αt In this case wf, the final angular velocity is zero, and α, the angular acceleration, is negative. For points on the edge of the wheel, the tangential velocity, v = Rw, and the tangential acceleration, a = Rα. The centripetal ...


0

You should know first the kinematics of rotational motion and the analogy with linear motion. Your stated equation for the velocity is used to describe a linear motion; However, the problem here is for a rotating body that requires you to know the relationship between the force applied on the disk and the torque which is related to the angular acceleration ...


2

As mentioned in the answer by sammy gerbil, there are two conservation laws (momentum and angular momentum) that are true, whatever the coordinate system, under specific conditions (no external force or torque). However, you are on to something with your intuitive view. If you study more advanced mechanics, you will learn about formulations of mechanics ...


0

There are only 2 laws of conservation here : angular momentum and linear momentum. Both are vectors. Vectors are expressed differently in Cartesian and cylindrical co-ordinate systems. In a Cartesian system the components of linear momentum are constant in the absence of external forces; in a cylindrical system the components can change (unless the object ...


1

I will use Einsteins-summation convention to make it more readable, this means I will sum over indices that occur twice within a product, i.e. e.g. $I_{ij} \Omega_j \equiv \sum_{j=1}^3 I_{ij} \Omega_j$. So, the way to get your equation is as follows: $\Omega_i \; I_{ji} \; \Omega_j \enspace = \enspace M_j \; \Omega_j \qquad \text{(1)}$ Now use: $\Omega \...


1

I do not know whether this is directly relevent to what you are seeing, but balls rolling in cylinders are known to have some weird behaviour. The famous and most infuriating example occurs in Golf where the ball goes down into the hole, rolls round a complete circuit and pops out again. (I.e in $\to$ out, rather than your out $\to$ in). There is a cute ...


1

They have solved the problem with respect to the center of mass(hence I=mr^2/2)as, though we know that there is friction acting on the system which is responsible for pure rolling it does not do any work because by definition of pure rolling the point remains instantaneously at rest. So while applying the work energy theorem we consider only the work done by ...


2

In the above question, they are analysing the situation from Centre of Mass. That's why they have taken the moment of inertia as $\frac{1}{2}mr^2$. However you can do the analysis from the instantaneous centre of rotation. The results obtained will be the same.


3

Your second disk cannot convert its rotational energy to gravitational potential energy. If I understand you correctly, both disks start out with some forward motion, and some rotation. Disk 1 will slow down as it travels uphill, converting both kinetic energy from its forward motion and kinetic energy from its rotation into gravitational potential energy (...


0

Cars can drive up hills. Their tires' rotation combines with friction to move forward. When the friction's removed, e.g. due to the road being icy, the tires' rotation doesn't have the same effect. That said, the textbook's a bit off: friction means that the disc's rotation matters, but this doesn't necessarily mean that it goes further as it could ...


1

Without friction, the second disk will roll, but not against the hill, thus it will lose kinetic energy exclusively in the form of translation - it's rotational kinetic energy will remain the same throughout the climbing, since the disk does not interact with the frictionless hill other than by normal force. The first disk will interact with the hill as it ...


19

Given that the disks started with the same kinetic energy while rolling at the base of the hill, shouldn't they reach the same height (i.e. same potential energy) as a result? First, let's assume that no slipping occurs on the hill with friction and that both disks are rolling without slipping before they get to their respective hills. You are forgetting ...


6

Given that the disks started with the same kinetic energy while rolling at the base of the hill, shouldn't they reach the same height (i.e. same potential energy) as a result? The height that each disc reaches will depend upon how much of its total initial kinetic energy (rotational + translational) is converted to gravitational potential energy. In ...


1

The way I would explain it is non-rigorous, but intuitively clear: The moment of inertia $MI$ of a mass $M$ with respect to an axis is closely approximated by $ MI = M R^2$, where R is the perpendicular distance from the axis to the center of mass of $M$. (This is very nearly true when the mass M occupies a small roughly spherical or cubical volume and R is ...


1

I believe that the Steiner principle can be used to show that axis through the centre of mass has the lowest moment of inertia (across the parallel axes), but it doesn't explain why. First, the principle is not intuitive. Second, to prove it you would use a construction similar to what is needed to show the statement in the first place. Here is my atempt to ...


0

Wikipedia answers this well (Rotation). A rotation is a circular movement of an object around a center (or point) of rotation. A three-dimensional object can always be rotated around an infinite number of imaginary lines called rotation axes (/ˈæksiːz/ AK-seez). If the axis passes through the body's center of mass, the body is said to rotate upon itself, ...


2

Although this question is rather vague I will try to comment on it. Circular motion is literally the motion of an object on a circle, subject to the centripetal force. For example when you swing around a ball at the end of a rope. For problems involving circular motion it is often enough to just know the velocity and mass of the object. Rotational motion ...


1

Orbital mechanics are weird. So, (hypothetically) we have the Earth travelling in a perfect circular orbit around the Sun. The radius of that orbit, the velocity of the earth, and the mass of the Sun are all fixed, and satisfy the relationship cited in the question. Now the Sun loses a small amount of mass. The Earth is still traveling at the same ...


1

Welcome to Physics.StackExchange! Let $M(t)$ be the solar mass and $m$ the Earth mass. Assuming circular orbit with radius $r(t)$, The gravitational force between them is $$ F = \frac{\gamma mM(t)}{r^2} = ma = \frac{mv^2}{r} \Rightarrow \sqrt{\frac{\gamma M(t)}{r}} = \frac{dr}{dt}, $$ where the orbital velocity is $ v = dr/dt$. Further assume the mass of ...


1

Consider Kepler's Third Law: $$T^2=\Big(\frac{4 \pi^2}{GM}\Big)r^3$$ where $T$ is the orbital period and $r$ the orbit radius (for a circular orbit). So: $$T^2 \propto r^3$$ where $\frac{4 \pi^2}{GM}$ is the proportionality constant. Since as it depends on $M$, the ratio $\frac{T^2}{r^3}$ will vary accordingly.


0

Note: the general momentum of inertia in 3D is a tensor, so you can't really divide by it. However, in flat motion, only component $J_{zz}$ matters, so it indeed works as scalar If the motion is limited to 2D, then $\pmb \omega=(0,0,\omega)^\intercal$ and $\mathbf J\pmb\omega = (0,0,J_{zz}\omega)$, which means that $\pmb\omega\times\mathbf J\pmb\omega=0$. ...


0

The three inequalities follow from $$I_{xx}=\sum_k m_k(y_k^2+z_k^2),$$ $$I_{yy}=\sum_k m_k(x_k^2+z_k^2),$$ and $$I_{zz}=\sum_k m_k(x_k^2+y_k^2).$$ They do not follow from the symmetry and positive-definiteness conditions that you mention. For example, the diagonal matrix with diagonal elements 1, 3, and 5 is symmetric and has positive-definite ...


0

1. The condition for rolling without slipping is $V_{cm}$=𝑅𝜔. 2. Irrespective of whether a body is slipping or not, velocity due to rotation of any point can be related as $V$=𝑅𝜔, where R is the radius of curvature and 𝜔 is the angular velocity. So yes, the relation holds good and the magnitude of tangential velocity can be found using the formula ...


1

So the $3\times 3$ Rotation Matrix R is a function of $\varphi$ and $\theta$ With Where equation (1) and (2) are the pendulum differential equation . With the solution of $\varphi(t)$ and $\theta(t)$ you can obtain the component of the angular velocity $\vec \omega$ equation (3). Example


Top 50 recent answers are included