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Solve by Conservation of Energy: $$K+U=\text{constant}\tag{1}$$ The kinetic energy has two components, a translational one and a rotational one: $$K=\frac12 Mv^2+\frac12 I\dot{\theta}^2$$ For slipping without sliding, $v=r\dot{\theta}$ The potential energy (gravity): $$U=MgR(1-\cos\theta)$$ Insert everything into $(1)$ and take the time derivative: $$\frac{\... 0 You can't conserve momentum here.The clamp of the pulley will provide an impulsive force.It is an easy question,try using the equation -"impulse=change i momentum" and it's rotational analogue too. 0 The other parts of the rod exerts a tangential force on the end which contributes to its downward acceleration. If you get a chance to watch a video of a tall brick chimney being demolished, notice that it breaks before hitting the ground. It is not designed to withstand the lateral forces required to match the angular acceleration of the upper end with ... 1 If both the masses after some time remains at rest ,and as all the forces here are conservative which means conservation of mechanical energy applies in this case So you see that in this case the gain in potential energy of m_1, would be equal to loss in potential energy of m_2 because ,if it doesn't do so then there would be a net change in potential ... 0 Yes, I believe you are correct. 0 In accelerating the car from rest, static friction DOES positive work which is the change in linear kinetic energy of the car. BUT it also does negative work on the wheel(it creates counter torque) which decreases the rotational kinetic energy of the wheel. So, the NET work done by static friction is 0. It simply converts some part of the rotational kinetic ... 0 The first equation is wrong. Angular momentum of a rigid body is first defined at the center of mass G as$$ \boldsymbol{H}_G = \mathbf{I}_G \boldsymbol{\omega} \tag{1}$$and then transferred to another point O with the following law$$ \boldsymbol{H}_O = \boldsymbol{H}_G + \boldsymbol{r}_{G/O} \times \boldsymbol{p} \tag{2} where \boldsymbol{p} is linear ... 0 Find the rotational and translational kinetic energy of a ball with rotational energy \frac{1}{2} I\omega_0^2 initially if put on a slant. \begin{align*} \frac{1}{2} m v^2+ \frac{1}{2} I\omega^2&=\frac{1}{2} I\omega_0^2\\ \frac{1}{2} m v^2+ \frac{1}{2} \alpha mR^2\omega^2&=\frac{1}{2} \alpha mR^2\omega_0^2&(\text{dimensions of }I\rightarrow [M]... 1 You are correct that before contact with the surface the wheel has KE=\frac{1}{2}I\omega^2 and after contact with the surface the wheel has KE=\frac{1}{2}I\omega^2+\frac{1}{2}mv^2. Since energy is conserved and since m and I are unchanged that implies that \omega must decrease in order to compensate for the linear KE. As the spinning wheel contacts ... 1 When you transfer velocity from one point to another in a rigid body you end up with an equation like \boldsymbol{v}_P = \boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{r}_P $$Acceleration is just the time derivative of the above with$$ \begin{aligned} \tfrac{\rm d}{{\rm d}t} \boldsymbol{v}_P &= \boldsymbol{a}_P \\ \tfrac{\rm d}{{\rm d}t}...

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That last term is the centripetal acceleration associated with the rotation of the rod.

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Because we need the minimum speed by which the circle gets completed i.e. it stops when the circle is just completed. If the speed given to $m$ is greater than $\displaystyle \sqrt{\frac{4gl}{3}}$ then it completes the circle as well as have some velocity at the lowest point. The torque of mass $2m$ is greater than that of $m$ about the center of rod and ...

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There is no answer that conserves energy. First, from elasticity, we know that the normal component of the velocity stays constant. We then have two unknowns: the new tangent velocity and the new rotation speed. Conservation of angular momentum around the point of contact (all forces act through it, so the net torque around it will be zero) provides one ...

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If the impulse is applied through the center of one of the spheres in a direction away from the other sphere, it will start with momentum ${m_1}{v_o}$ and kinetic energy (½)${m_1}{v_o^2}$. The spheres will oscillate on the spring. Later, if the surface is friction-less, at maximum extension of the spring, the two spheres will have the the same velocity, ...

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Your CM is revolving around a given axis in a circular motion, but each particle isn't just doing a circular motion, so you can't apply that formula for its velocity. The particles are moving around the CM in some arbitrary fashion. The velocity of each particle, as seen by your observer $O$, will be $\vec{v}_i=\vec{V}^{}_{CM}+\vec{v}_i'$, where $\vec{V}^{}_{... 1 It can if the shaft is not part of the fan assembly. Consider the equal and opposite forces produced by the spring. If they act on the same body (the shaft is part of the fan) then nothing happens because it is an internal force. Also since the shaft does not rotate relative to the fan, the string will never get unwound. But if the shaft is grounded, and ... 0 All great and valid answers here, and they have helped me see the key piece of the puzzle I was missing: the subtle point that although the instantaneous axis of rotation has zero velocity, it is accelerating upward (centripetal acceleration). Therefore to apply Newton's second law (or Euler's law) about this axis, we need to take into consideration the ... 2 To calculate the equation of motion we obtain the sum of the torques about point A, because we don't have to take care about the contact force. first I obtain the vector u from point B to A $$\vec{u}=R\,\begin{bmatrix} 0 \\ -1 \\ 0 \\ \end{bmatrix}- R\,\begin{bmatrix} \sin{\theta} \\ \cos(\theta) \\ 0 \\ \end{bmatrix}=-R\,\begin{bmatrix} \... 2 In general, you can only apply \tau_C = \tfrac{\rm d}{{\rm d}t} L_C about the center of mass C. The expression about a different point is quite more complex. You can see that taking the torque about another point A (not the center of mass C), and the derivative of angular momentum about A isn't enough to solve the problem. Using the standard ... 4 The answer is that the moment of inertia is changing not only due to the instantaneous rotation about the contact point, but also because of the horizontal motion of the cylinder. The position of the rod (I won't write the z component of vectors) is$$r=R(\sin \theta, 1+\cos\theta)+\int_{t_0}^t (R\omega(t'),0)dt'$$where at t=t_0, the cylinder is above the ... 21 I don't know if there's a beautiful solution for this. I'd love to see it, if it exists. What I can do is show you how I slogged my way through it. All praise to the mighty Mathematica. Part I: Obtaining the Equations of Motion First, we can dispense with the cylinder and rod and consider only a point mass M on a ring of mass m and radius R. Define ... 1 It seems I've figured out why the tube will jump if the mass of the rod is large enough, but I can't calculate the exact threshold. My proof of the jump is below. Let's suppose that the mass of the original tube (i.e., without the rod) is infinitesimal, whilst the mass of the rod is finite. Let's also assume that the tube won't jump. I'm going to prove the ... 11 Turns out the tube does jump if the rod is no less than 13 times the mass of the tube. My previous answer had a couple of mistakes that yielded the wrong result, here is the updated one. Let M be the mass of the tube, m the mass of the rod and R the radius of the tube. Let \theta be the angle between the vertical and the direction of the rod from the ... 2 Parallel axis theorem is for the rotation of a body about a single axis that does not pass through the center of mass. But the earth's motion cannot be simply described in this way in the frame where the sun is at rest. It is more usually described as a combination of rotation and orbital movement, where each of these motions has different angular speeds. ... 1 Torque on earth due to sun is 0 as gravitational force always acts towards center. In the formula of torque of torque which is \tau = \ rF sin\theta. Now \theta becomes 0 as Force and r vector acts on same line (r is perpendicular distance from the force F)and hence overall torque becomes 0. As net torque is 0 about earth's center, therefore angular ... 3 Look at this image of the earth. (image from Gravity for Geodesy I: Foundations) The gravitational force (red arrow) points to the center of the earth. The centrifugal force (yellow arrow) points away from the axis of rotation. The total force (green arrow) is the vector sum of the two above. In general (except at the poles or at the equator) it does not ... 0 Point mass in circular path is translational and kinetic energy is \frac{mv^2}{2} Though, you can write it in terms of its moment of inertia and angular speed as \frac{I{\omega}^2}{2}. But the two expressions do not represent kinetic energies of two different motions. They are one and same so don't add them, either you look from center or from somewhere ... 0 Suppose i have a point mass which is moving in a circular path Let the mass of the particle be m If we see this particle from a frame of reference located not at the centre of the circle but say somewhere else from where the motion appears to be only translational. And the velocity of the body with respect to this inertial frame(another assumption) is \... 1 Let's call the falling direction z (-z to be exact). The rotation of the ball causes a (small) force perpenticular to the moving direction (so some combination of the x and y direction), due to the difference speed of the air flow to the sides of the ball Magnus effect. As soon as the ball gets velocity components in x or y direction, the force ... 1 I think you are over-complicating the problem by using the Lagrangian. I'll give you a few pointers that should allow you to do the maths yourself. Use conservation of energy:$$ K + U = E = \mathrm{constant}, $$where K is the kinetic and U the potential energy. U = mga(1-\sin\theta), so that at \theta = 0, with the top upright, the (gravitational) ... 1 Following the suggestion of @maverick I solved the problem using the work-energy theorem using$$\frac{1}{2}mv^2 = \int_{0}^{R} m\omega^2 r \; \rm{d}r\implies \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2 R^2 \\ \implies v = \omega R $$0 If you could ignore direction, this would just be a constant acceleration for a known time. 0 The satellite will indeed accelerate, but instead of falling to earth, it will form a thinner and longer orbit. And concerning the acceleration on earth, it is indeed changing due to centripetal acceleration, and the change is the smallest on the poles. 0 According to classical mechanic when you set anything on a "spinning" motion it's required a centripetal force, acting on the rotating body, in order to maintain the body on its curved path; otherwise the body will just start moving in a straight line. In this case we are dealing with a fluid and not a solid object, but regardless the same ... 1 If would like to elaborate on the answer give by Leo L. Tidal forces have to be considered, too. Those will eventually lead to a rotation of the coin locked to its orbit, like moon. In this specific set up, however, I reckon that tidal forces are so small that they can be ignored for all practical purposes. The term „practical purpose“ here includes the moon ... 0 "How will the cone move when it starts motion due to its weight" to answer this equation you have to obtain the equations of motion. you have 3 generalized coordinates rotate about then z -axis angle \psi incline plane x position s_x incline plane y position s_y thus you get : Kinetic energy$$T=\frac 1 2\,{{\it \dot{s}_x}}^{2}m+\frac 1 2\,{... 1 If you are dealing with water, then it will behave nearly like an inviscid fluid, meaning that, aside from a very thin boundary layer near the wall, the fluid will not be rotating as the can rolls down the incline. Basically treating the water as inviscid is equivalent to allowing the water to slip at the wall. This would prevent the can rotation from ... 29 Yes. Paint half (as in a filled semicircle) a face of the coin black, while leaving the other half shiny. Place the painted face parallel to the Sun's surface. The two sides of that face will now experience different radiation pressures from the solar radiation: in particular, the still-shiny half will experience greater radiation pressure because it ... 5 Assume there is no other force (such as friction), and say the coin is near the Earth and experiences the Earth's gravitational field, then with classical mechanics, the gravitational force only acts on the center of mass of the coin, and produce no torque, so the coin, if it is spinning initially, will remain spinning forever. This is the rotational analogy ... 0 If you are accelerating in your car, then the friction force from the road acting in the forward direction on the bottom of the wheels is moving forward with your car. As the only net external force, it is doing the work to accelerate the car. If you apply the breaks, the friction force acts backwards, moves with the car, and does negative work to remove the ... 0 Centrifugal force depends entirely on the chosen reference frame. It's consequences do not depend on the mass of an object (which cancels out from any equations). If you choose a reference frame in which the ball is stationary, there will be a centrifugal force, balanced by a normal force from the cylinder, and you only need this to be enough to generate a ... 1 because as we know when relative velocity is zero , friction shouldn't act. This is false. Static friction acts when the relative velocity between surfaces is$0$. For a simple example, take a heavy object and start pushing on it without it moving. Static friction is the force that opposes your applied force before the object starts moving. Because of this, ... 1 This problem is a little tricky in a way that probably was not intended. It would make more sense if the string was replaced with a spring. When you drop the mass, it accelerates. But the disc stays still. When the string becomes taut, that must suddenly change. The mass and the edge of the disc suddenly start moving at the same speed. If they were connected ... 0 The torque is always horizontal (in the direction of precession). Since$\vec{\tau}=d\vec{L}/dt$, the vertical component of$\vec{L}$does not change. However you are right in your intuition that the torque doesn't only affect the horizontal component of the "spin (figure) axis" of the top. For a fast top, this axis is very closely, but not ... 0 The general formula for a rigid linear molecule is $$E_{r}=\frac12 I \omega^2$$ where$I$is the tensor of inertia. In the particular case of a diatomic molecule the inertia tensor is, as you write$I=\mu R^2$. So, if you have a triatomic molecule you just have to use its inertia moment (calculated with respect to the rotation axis) in the general formula. 0 Does that mean... Moment of force = work done by force No. In the case of angular motion (motion produced by torque, or as you say, moment) force is replaced by torque and linear displacement is replaced by angular displacement. Or, $$W=\int \overrightarrow F.d\overrightarrow x$$ Is replaced by $$W=\int \overrightarrow τ.d\overrightarrowθ$$ Where$θ$is the ... 3 You can try to solve this problem using law of conservation of energy and using law of conservation of angular momentum and you will get different answers. The thing is that you can't use conservation of energy law in this case. At the moment when the string became taut some kind of inelastic impact would happen and some portion of energy would lost. ... 0$\omega$can take on any value as long as it is large enough to allow the strings to not have any slack in them. Yes, in the case of$l\omega^2=\sqrt 2g$we would have$T_2=0$. For$\omega$smaller than this we would get slack in the bottom string as the mass would come closer to the pole. For$\omega$larger then this we would just get larger tensions. ... 1 Notice,$V=R\omega$is the linear or tangential velocity of any particle of a body rotating at angular speed$\omega$at a particular instant of time$t$because the tangential velocity$V=R\omega$keeps on changing in direction from instant to instant while the magnitude$R\omega$is constant if angular velocity$\omega\$ of rotating body is constant. ...

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Look neglecting friction makes it simple. We have to think only about the applied force for rotational motion as well as translational motion. So since there is an external force , the body will have some translational motion in the forward direction. This will happen for sure and we can find the acceleration using F=ma Now we have to look for the rotational ...

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