# Tag Info

1

If what you mean is Center of Mass, then, no, it will not change the angular velocity (if there are no other external torque acting on the object, and the object is not attached in any way to other objects, i.e, glued, or attached to a fixed rotating rod or another object, who's rotating axis does not intersect the Center of Mass of the object attached). In ...

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Point C is sure instant centre of rotation for both gears, otherwise they would get teeth broken if any relative slide to each other. Analogy is a wheel on a road having instant centre of rotation at the bottom point thus velocity of top point is twice more than of the car. As far as I understand, confusion point is that first you calculate VO in respect ...

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If the center of mass is the same as the center of gravity then it does not change the angular velocity. As far as we know the inertial mass is the same as gravitational mass and hence the center of mass is always also a center of gravity (defined as a center of gravitational mass). Some people define a center of gravity as point where the average ...

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It would be easier to answer your question clearly with a drawing. In the following, the angle coordinate of the pendulum is the angle it makes with the vertical line. When the pendulum swings right(left), the angle will be positive(negative). With this setting, I get the exact same answer as you by working out the equations of motion. However, there ...

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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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It's because of the product rule of derivatives, that states d(fg(t))/dt=f(t)(dg/dt)+g(t)(df/dt). In this case, let's call this part: (-isinθ+jcosθ)=g(t), and solve it: d(rωg)/dt=r(dωg/dt)=r(ω*(dg/dt)+g(dω/dt)). If you substitute g for (-i*sinθ+jcosθ), you get the equation for acceleration your book presents.

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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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It is vector because it has a magnitude and a direction. Also typically the components of $\vec{\omega}$ are evaluated based on an inetial coordinate frame and thus only represent the motion of the body and not of the measuring frame. Rotational speed $\omega = \| \vec{\omega} \|$ Direction of rotation $\hat{z} = \frac{\vec{\omega}}{\omega}$ ...

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Because that point is the instantaneous velocity center i.e. its velocity is zero. But, it doesn't mean that its acceleration is zero too. If you want to calculate the acceleration of a point like this method, you had to find the instantaneous zero acceleration center.

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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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REVISED ANSWER : The tension in the string will become zero when $v^2/r = gsin\theta$. v is still +ve when this happens. v will only become zero later. So you are correct : first T becomes zero (which cannot happen until $\theta > 0$), then the stone moves as a projectile on a parabolic path, then v beomes zero.

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The torque derivation is as follows: $$\sum \vec \tau = \frac{\mathbb{d}L}{\mathbb{d}t}$$ The magnitude of the torque (measuring $\theta$ with respect to the horizontal) is: $$|\vec{\tau}| = mgl\cos\theta$$ The angular momentum $L$ is given by: $$L = mr^2\omega = ml^2\frac{\mathbb{d}\theta}{\mathbb{d}t}$$ So the rate of change of angular momentum ...

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The first green part is the Rodrigue's rotation formula. The second green part is a small angle approximation for $\delta \theta$.

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If a body of mass m hanged on a string is moving, let uniformly, on a circle fixed relatively to the ground, then an observer G on the ground uses the 2nd Newton Law : $$\mathbf{F}=m\cdot \mathbf{a} \tag{01}$$ and finds the relation between the force $\mathbf{F}$ and the acceleration $\mathbf{a}$. For observer G there exists a "real" force, the tension ...

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Note that you have to swing the pail with a certain minimum speed for the water to stay in. That minimum speed is such that when the pail is at the top of the arc, the rope accelerates the pail downward faster than gravity accelerates the water downward. Otherwise, the water falls out.

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If water particles move in a circular path its because of some net force towards the centre. This net force is usually called "centripetal force". Don't ever put centrifugal force into the description. It is not a force, but just a name of the "feeling" that your body (or in this case water particles) want to move out of the circular motion but can't.

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The loss of energy (efficiency<1) manifests itself in two ways Friction. When doing free body diagrams add frictional forces at the sliding contacts and find which coefficient of friction $\mu$ gives you the efficiency values you expect. Structural damping (hysteresis). This is more difficult to put in the equations of motion, because they assume ...

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First of all, the transmission of force between any two pairs of mating gears happens due to surface contact between the two and hence through friction. Next, any loss is a dissipation in energy. Though I am not that well-versed in vibration analysis, I know that an periodic dissipation of energy is accounted in the force-equation using a damping ...

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Newtons 2nd law only holds in an inertial frame. If you use the center of mass of the rolling object as your frame of reference then the frame will be accelerating and F=MA won't hold. However if you fix the frame of reference to a relatively stationary point, say the surface of the earth, it will. Even in this case the frame is not truly inertial, but ...

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