# Tag Info

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I have managed to figure out what was going wrong. My mistake was not realising that the spool would have to be at the side of the string, which gives us the constraint: $\dot{y}=r\dot{\theta}$, we therefore have our Lagrangian: $$\mathcal{L}=\frac{1}{2}m\dot{y}^{2}+\frac{1}{4}mr^{2}\dot{\theta}^{2}+mgy=\frac{3}{4}mr^{2}\dot{\theta}^{2}+mgr\theta$$ ...

0

In the experiment (ii) there is a mass $m/2$ , right? but in the (i) experiment there is only one mass of $m$ on one side of the spring, yes? but how about the other side? is there any force exerted on the spring? well, the answer is yes. according to Newton's third law, there is a reaction force acting on the upper side of the spring that has the same ...

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You posed two questions: First, Earth exerts a gravitational force on the car. The car, in turn, exerts a gravitational force on Earth. These are equal and opposite. This can be seen from the fact that the formula for magnitude of the (Newtonian) gravitational force: $$F_g=\frac{GM_1M_2}{r^2}$$ remains of exactly the same form if the masses are switched. The ...

-3

They have to meet at the same time to keep the overall pressure behind the wing equal. The velocity of the air on the top run relatively faster because it has a greater distance to travel (due to the curvature of the wing) and all in the same time as the air of the bottom. Why the same time, because when the air divides it still has to keep pressure ...

0

If the man pulls a length $L$ of rope through his hands, both he and the counterweight will rise by $L/2$. The rope is assumed to be massless, and hence the force of tension is uniform throughout it at all times. As the man does work to pull the rope through his hands, he increases the magnitude of the tension force in the rope to greater than his weight. ...

1

You have a couple of mistakes. First, if you say that $G = 6\times 10^{-11} k$, then $k$ should be $\frac{\text{m}^3}{\text{kg}\cdot \text{s}^2}$. If we instead define $k$ as you did, then it is a dimensionless number: the conversion factor between the two sets of units. You should rather have said that $6\times 10^{-11} / k$ is the value of $G$ in the ...

3

Perhaps a better question to ask is: why is a single pendulum non-chaotic? Almost all real systems are chaotic at least to some extent; the fact that we can write out the solution for a single pendulum for all points in time is really quite peculiar, and only true because it is a highly simplified system. The reason these non-chaotic systems are so prevalent ...

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Chaotic is not the same as random. A chaotic system is entirely deterministic, while a random system is entirely non-deterministic. Chaotic means that infinitesimally close initial conditions lead to arbitrarily large divergences as the system evolves. But it's impossible, practically speaking, to reproduce the same initial conditions twice. Given ...

2

The force due to gravity balances the buoyant force exerted on the block The buoyant force is there because of gravity (There is a difference in pressure as we go deeper in to the ocean). There's an easy way to think it through. Imagine a beaker with some water, and it is standing still. There is no internal motion. Consider a small portion of this ...

3

The buoyant block does exert a force on the water, it's force is equal to the mass of the displaced water, so the pressure of the water immediately beneath the block is exactly the same as the pressure of the water at that height in the rest of the container. Indeed, the mass of the system is just the mass of container with water + mass of block

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Hint:Another possible way to solve it would be to observe that $$a=\frac{dv}{dt}=\frac{dv}{dx}*\frac{dx}{dt}$$ Hence $$a=v\frac{dv}{dx}$$ Now according to question power is constant Hence P=k(say) $$Fv=k$$ $$\Rightarrow mav=k$$ $$\Rightarrow mv^2\frac{dv}{dx}=k$$ Solve the differential equations with the given limits to get the an equation of v in terms of ...

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Hint : You assumed the force to be constant by using $W=Fx$ which is wrong. It is $W=\int Fdx$ Use $P=\frac{dW}{dt}$ Use work energy theorem. Use calculus. Find distance covered when speed is $6ms^{-1}$. That should eliminate your variables if the question has sufficient information.

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What you need is to match mechanical impedance between what the motor produces, and the tuned mass damper. Read more here http://www.bksv.com/doc/17-179.pdf and here http://arkansas.s.jniosh.go.jp/en/indu_hel/pdf/43-3-3.pdf

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Read the blue box in NCERT on page no 219 with the heading "ELECTROMAGNETIC DAMPING". If we do the experiment in a aluminium pipe, then due to eddy curretns the glass bob will reach first as the motion of spherical bob is opposed. BUT if we do the experiment in a PVC/Plastic pipe or in air...both will reach at same time.

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Well basically what you said is true, the maximum height, in this case, depends on the initial velocity and the angle $\theta$. So if you consider that the maximum height to be 2R, and by using the trajectory equation, while replacing $x_{y_{max}}=\frac{Total \;Distance}{2}=f(u)$, you'll get $u=f(R,\theta)$. Now I did the calculation and I got ...

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The value of initial velocity will be different for different angles θ with the horizontal.. So I got this result. $$u=(gR/(sinθcosθ-cos^{2}θ))^{1/2}$$ or $$u=(2gR/(sin2θ-2cos^{2}θ))^{1/2}$$ or $$u=(39.2/(sin2θ-2cos^{2}θ))^{1/2}$$ This is my attempt for the solution(i have attached image): From A to B displacement is FB From C to B displacement is ...

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Knowing only the force is not sufficient to calculate the original height. This is because in order to find this, one needs to calculate the velocity or momentum of the object at the instant before it collides with the ground. Consider how the motion of the object is affected as it collides with the ground. As soon as the object makes contact, the force of ...

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Force alone will give you no information on the height the object fell from. Force is mass times acceleration, the latter is given by $g=9.81m/s^2$, which is (roughly) a constant on the earth's surface. It does not depend on the distance the object travelled before it hit the ground. The information you need is momentum, which contains the object's velocity. ...

2

Let's suppose I have some system and I know $M$, the system's total mass, $\vec{r}_{cm}$, the system's center of mass position and $\vec{L}_{cm}$, the systems angular velocity in the frame where the center of mass is the origin. How do I find $\vec{L}'$, the angular momentum with respect to some other origin, say $\vec{r}_{cm} + \Delta \vec{r}$, which is ...

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HINT: Use $H_{max}=\frac{u^2sin^2\theta}{2g}$ (max height) and $R=\frac{u^2sin2\theta}g$ (Range).

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The friction acts only when the contact point slips relative to ground. You can consider speed of lowest point to be sum of $v$ and $\omega r$ with proper directions. Friction acts till there is slipping and condition for no slipping is $v=\omega r$ when v is right and $\omega$ is clockwise acc. to diagram As in first case, the lowest point in always ...

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1st que- as the point in contact of the disk and the plane is not moving, no work is done. and as for the second question; no body is perfectly rigid, therefore due to deformation there is loss in energy. that is why the disk stops.

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You can no longer consider the ice block as a system once it starts melting, so it no longer remains a rigid body. Consequently, you cannot apply Newton's law on it. Peace ;)

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why wouldn't that straight line be in the direction of acceleration why do you think the acceleration line be in the direction of tangent? the tangent is where a body would have kept moving if the rope didn't pull it. so the vector of speed changes towards... where the rope is attached, i.e. perpendicularly. acceleration is the change in velocity ...

2

Frictional force opposes sliding motion, basically. Car tires produce centripetal force by changing their angle relative to the rest of the car's orientation. The tires do not slide in the direction of the tires' orientation: they roll. Friction in this direction rotates the tires, or if the engine is applying force to the wheels during the turn, friction ...

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The frictional force is opposing the tendency for the car to slide off the road. Think of a merry go round: If it is really fast you will struggle to stay on. If it goes really fast you will grab on to the bar and your body will point radially outward from the center of rotation. Eventually you be able to hang on and you will fly off. This is the same idea ...

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i think that both the bobs will reach the ground at same time because mass of both are nearly the same as they are of equal sizes and consideration of earth's magnetic field is not very effective in my point of view.

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Listen guys..I don't think we need to just work out that much on the problem. This is CBSE and it doesn't require that much use of brain...and it also won't go much out of syllabus! I think that the question demands us to think of earth as a magnet...a bar magnet...and a magnet would at any case attract metallic bob...like a bar magnet attracts metallic ...

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While I agree with the caveats made by dmckee in his comments, there is an obvious interpretation of stopping power as the change in momentum caused by the projectile. The mass and velocity of the projectile are $m$ and $v$ respectively, and the mass of the target is $M$. Since the target is stationary the initial momentum is just $mv$. Assuming the ...

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No, these building are still tiny compared to earth's crust mass distribution. One would need to build whole mountain ranges to detect changes in earth gravity field with high precision instruments. And even those wouldn't changed earth orbit measurably because even a mountain range is tiny compared to the mass of the whole earth. However mountain ranges ...

2

The fly would move out of the way. The fly (when right in front of the pilots face) has the same forward velocity as the plane. If the airplane accelerated forward then the fly would "fly backwards" (no pun intended) just as you think you do in a car that accelerates quickly. However, this is actually not you flying backwards, it is simply the car going ...

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I haven't checked the arithmetic, but the method to calculate the acceleration is right. The easiest way (to my mind) to find the tension (T) in the string is to apply F=ma to the smaller mass block. Hope fully if you do that, you will also find that (T-6.4) will also accelerate the larger mass at the required rate as well.

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The tension in the string has nothing to do with gravity. Just consider the forces acting in the horizontal direction for each mass separatedly.

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As the blocks are moving horizontally, the gravity and normal forces cancel each other. So your logic about finding the tension is faulty. You should use Newtons second law on each separate block. To find the tension you could suffice with the last one. If there is friction you could add the friction force to the equation, and think about what this would ...

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Emf on a conducting object induces eddy currents. These in turn decay due to the electrical resistance of the object. What you end up with is energy in the form of heat. When you compare the two objects (essentially a conductor versus a non-conductor), a portion of the potential gravitational energy goes into generating eddy currents. That means the ...

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metal ball would hit the ground first. there wouldn't be any effect of induce emf or magnetic field. it only depends on gravitational force, as F=mg the mass of metal ball is greater, therefore, it will reach faster than glass ball, whose mass is less than metal ball.

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Yes the skater does increase the angular momentum by doing work; pulling her arms in. You do work on a swing (sitting up and down) to increase your angular momentum likewise.

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Think about the density of Lithium which is lower than the density of glass. The lithium ball many hit the ground later than the more dense glass due to air resistance. The experiment requires a procedure to put coating on the lithium ball to avoid it from burning in air.

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It is the elimination of friction on the ground. The friction on the air is very small, as is the resistance of the rope to twisting. No matter how smooth the floor, the friction will be much higher than the resistance of the hanging weight. This is why air bearings were invented.

-1

Lets see, the engine applies torque (pistons or not it does not matter) and if an equal and opposite reaction from the air is applied to the propeller then the engine speed is constant. In general, the torque imbalance between the engine and the drag torque results in engine acceleration or deceleration. And that is the story about torques. If you are ...

1

When you solve a problem like this, you are using a system of reference (actually you use one in all problems, but here it is very explicit). In this case, the easiest one is y in the vertical and x in the horizontal. Almost all the forces are already in one of these 2 directions. Namely, you have all the weights pointing downwards, so in the -y direction ...

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The first equation is a very close approximation since m (satellite's mass) << M (Earth's mass) so m can be ignored. The second equation is the mathematically correct one.

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The motion is in two dimensions. Vertically it is a decelerated/accelerated motion. Horizontally the velocity is constant. So split the initial velocity in a vertical and horizontal component. You calculate the time for traveling up and down using the equations you already posted. With $x=v_x t$ you can calculate the horizontal distance.

1

It will in general depend on the shape of the object. If it has a large concentration of mass at the edge you are lifting then the force will be close to its weight; if its mass is concentrated near the other edge then it will be very small. The general case is solved using the law of levers: If $d$ is the distance from the fulcrum to the centre of mass ...

1

If you are lifting on one edge and it is resting on the other edge, and the edges are an equal distance from the center of mass, then the answer is $$\boxed{F = \frac{1}{2} W}$$ If you are lifting with a distance of $\ell_1$ from the center, and the pivot is $\ell_2$ from the center then $$\boxed{ F = \frac{\ell_2}{\ell_1+\ell_2} W }$$ This is commonly ...

2

Surprisingly, the answer is that yes you do, though the effect is very small. To see this consider the following (highly exaggerated) diagram of the lift shaft: The Earth rotates at a constant angular velocity of one rotation every 24 hours ($\omega = 7.27 \times 10^{-5}$ radians/sec). The tangential velocity of a part of the lift shaft at a distance $r$ ...

0

He may experience a Coriolis force, but that is very small in magnitude. I am not sure if you could measure it. The Coriolis force, however, is only experienced by observers in a moving coordinate system when moving relative to the moving frame of reference. As you situate your elevator on earth, we have a rotating coordinate system, that rotates with the ...

1

A well executed barrel roll maintains the force balance you experience at rest with "gravity" oriented in the direction you experience as "down" (that is the direction from your head to your feet) due to centripetal acceleration. If you weren't looking outside, you might not realize the roll even took place (if the pilot is good). For those not convinced ...

1

I am not sure what you meant by: "I figured I could simply calculate the magnitude of the components since that will give me the distance" But the idea is use the kinematics equations for x and y: $x(t)=x_{0}+v_{x0}t+1/2at^2$ and $y(t)=y_{0}+v_{y0}t+1/2at^2$ These equations are derived from integrating the acceleration function ...

1

"For every action, there is an equal and opposite reaction" This means, forces exist as pairs. When there is an interaction between $A$ and $B$, an action-reaction pair between them is produced. Which one is action and which one is reaction depends on your frame of reference. Now to the static friction. Your question is pretty vague so I'm going to ...

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