# Tag Info

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The object will continue accelerating to just below light speed (speed limit of the universe), but only in a vacuum. Unfortunately I only know a little about air resistance, but apparently the faster an object travels the more the atmosphere tries to resist the object. This is where aerodynamics comes in. I'm sure you're familiar with the classic image of ...

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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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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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Note: Solution attempts to problems like these are susceptible to sign-flipping errors, so we must ensure clarity about what positive and negative values mean. Definitions. Let $N_1$, $N_2$, $f_1$ denote magnitudes only, as requested in the problem statement. As a consequence, the fact that the forces quantified by $f_1$ and $N_2$ are supposed to counteract ...

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You need to show some work. Draw a free body diagram. Then use Newton's laws with $\sum F=0$ (since you don't want the child to move i.e. accelerate) in both the x and y directions and solve the two equations for the unknowns. There are three forces: friction, weight ($mg$), and normal force.

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Sorry to contribute almost 2 years after question was asked, but I can't help it the Google brings me to this discussion, right now ! ... ;-) The question is strange or weird, in the way that asker start presuming something about "The magnitude of acceleration", as if with the moving legs, one must make abstraction of the whole body ? Accelleration & ...

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If the bumpers where vertical then the contact point would be at the center and since gravity is more than the bounce force it means there isn't going to be enough friction in change the rotation of the ball when the direction of the ball changes. If the contact point is further up, then the contact force is towards the center of the ball, and hence is ...

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Your second therm is still referring to the conservative force acting on the object. For an ideal spring this would mean: $\omega_0^2=\frac{k}{m}$. However in the case of a damped oscillator there will be both conservative and not conservative forces acting on the object with mass $m$. But this does not mean than the potential energy of the conservative ...

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I assume your car is front wheel drive. The phenomenon is simply Newton's third law in disguise. The car exerts a torque on its forward axle and the wheels exert the same magnitude, opposite sense torque on the car. Normally, the torque is not so big, because as soon as it is exerted on the wheels by the car, the wheels push backward on the road and the ...

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First of all don't insert actual numbers until the end. It makes it much easier to keep track and check whether your units check out. This problem is easier if you invoke conservation of energy. Simply equate: At t = 0. -Potential gravitational energy. At the end. -Kinetic energy of the toolbox. -Dissipated energy due to friction. You will find that ...

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Yes, that equation will still give the correct value for the energy of the oscillator system at any point in time, assuming of course that you know dx/dt and x at that time. If there is an external dissipative force on the system (damping) you will find that the value of E decreases with time. But the energy of the oscillator itself is still the sum of ...

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This problem is an example of rolling without slipping. A very good explanation of this concept is given here. In this case, it implies that rolling without slipping occurs if $\tan \theta \leq 3 \mu$. The expression validates one's intuition too. Its easy to observe that a cylinder tends to roll without slipping when kept on a wedge with lesser slope.

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