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Consider a small frictionless puck perched at the top of a fixed sphere of radius \(R\). If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy to find the puck's speed as a function of its height, then use Newton's second law to find the normal force of the sphere on the puck. At what value of this normal force does the puck leave the sphere?]

Short Answer

Expert verified
The puck descends a vertical height of \(\frac{R}{3}\) before leaving the sphere.

Step by step solution

01

Apply Conservation of Energy

At the top of the sphere, the puck has potential energy and no kinetic energy. If the puck descends to a height, say, \(h\) from the top, it will lose some potential energy and gain kinetic energy. Set the zero for potential energy at the center of the sphere when the puck is at height \(h\), then:\[mgR = mg(R-h) + \frac{1}{2}mv^2\]Simplifying the equation gives:\[v^2 = 2gh\]Calculate the speed \(v\) in terms of the height \(h\):\[v = \sqrt{2gh}\]
02

Use Newton's Second Law

As the puck slides down, it moves on a circular path, requiring centripetal force, provided by the normal force \(N\) and gravity component. Applying Newton's second law in the radial direction:\[N + mg\cos(\theta) = \frac{mv^2}{R}\]Substitute for \(v^2\) from the energy equation:\[N + mg\cos(\theta) = \frac{m(2gh)}{R}\]Solve for the normal force \(N\):\[N = \frac{m(2gh)}{R} - mg\cos(\theta)\]
03

Determine the Condition for Leaving the Surface

The puck will leave the surface when the normal force \(N\) becomes zero. Therefore, set \(N = 0\):\[0 = \frac{m(2gh)}{R} - mg\cos(\theta)\]This simplifies to:\[2gh = gR\cos(\theta)\]Simplify this to find \(h\) in terms of \(R\):\[2h = R\cos(\theta)\]Now, relate \(\cos(\theta)\) with \(h\) using the geometry of the sphere:\[h = R(1-\cos(\theta))\]Substitute \(h = R(1-\cos(\theta))\) into the energy equation above:\[2R(1-\cos(\theta)) = R\cos(\theta)\]Solving the equation gives \(\cos(\theta) = \frac{2}{3}\), so:\[h = R(1 - \frac{2}{3}) = \frac{R}{3}\]
04

Conclusion

Therefore, the vertical height that the puck descends before it leaves the sphere is \(\frac{R}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton's Second Law of motion is one of the cornerstones of classical mechanics, describing how the motion of an object changes when subjected to external forces. It is succinctly expressed by the equation \( F = ma \), where \( F \) is the force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration produced.
This law is particularly useful when analyzing situations where objects follow a curved path, like our puck sliding down the sphere. The law tells us that the net force acting on the puck—not only determines its acceleration, but also dictates whether it stays on the sphere’s surface. In our case, the force causing this centripetal acceleration is provided by the normal force combined with the component of the gravitational force that points towards the center of the sphere.
Here’s how it worked in the exercise: As the puck slides, its weight component \( mg\cos(\theta) \) and the normal force \( N \) together supply the centripetal force necessary. Substituting for the puck’s speed from the kinetic energy relation, you find that the normal force varies with height. The critical point is when \( N = 0 \) which indicates the puck loses contact with the surface. Thus, using Newton's Second Law, you find the condition leading to the puck’s departure from the sphere.
Potential Energy
Potential energy is the energy an object possesses due to its position in a force field, commonly a gravitational field on Earth. In our scenario, the potential energy (PE) of the puck is highest at the top of the sphere due to its elevated position and zero kinetic energy at rest.
The exercise requires us to apply the principle of conservation of energy, which states that the total energy in an isolated system remains constant. This principle guides us to understand how potential energy converts into kinetic energy as the puck slides without friction. Initially, the puck’s energy is purely potential since it’s stationary, indicated as \( mgR \). As the puck descends to a height \( h \), its potential energy is \( mg(R-h) \). The converted energy becomes its kinetic energy \( \frac{1}{2}mv^2 \).
This transition between potential and kinetic energy is vital when calculating the speed \( v \) in terms of the height \( h \), reaching \( v = \sqrt{2gh} \). Understanding potential energy's role in this dynamic helps explain the puck's behavior as it moves and eventually detaches from the sphere.
Centripetal Force
Centripetal force is essential in understanding the motion of an object that travels along a curved path, like a puck sliding down a sphere. It's the inward force that maintains the object's curved trajectory, pulling it towards the center around which it moves.
In the sphere scenario, as the puck slides down, it traces a circular path, creating the need for this compelling inward force. Remember, centripetal force itself isn't an independent force; rather, it results from components of other forces, such as tension, gravity, or normal force.
For the puck, the necessary centripetal force is the sum of the normal force from the sphere and the inward component of gravity, \( mg\cos(\theta) \). When calculating if and when the puck will leave the sphere's surface, issue raised was when centripetal force couldn’t be maintained because the supporting normal force equaled zero. At this juncture, the puck can no longer maintain its track along the sphere, ending up detaching at the height \( h = \frac{R}{3} \). Therefore, centripetal force plays a critical role in such analyses, ensuring we understand the conditions under which an object stays on its curved path.

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Most popular questions from this chapter

Consider a head-on elastic collision between two particles. (since the collision is head-on, the motion is confined to a single straight line and is therefore one-dimensional.) Prove that the relative velocity after the collision is equal and opposite to that before. That is, \(v_{1}-v_{2}=-\left(v_{1}^{\prime}-v_{2}^{\prime}\right),\) where \(v_{1}\) and \(v_{2}\) are the initial velocities and \(v_{1}^{\prime}\) and \(v_{2}^{\prime}\) the corresponding final velocities.

(a) Consider an electron (charge \(-e\) and mass \(m\) ) in a circular orbit of radius \(r\) around a fixed proton (charge \(+e\) ). Remembering that the inward Coulomb force \(k e^{2} / r^{2}\) is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to \(-\frac{1}{2}\) times its \(\mathrm{PE}\); that is, \(T=-\frac{1}{2} U\) and hence \(E=\frac{1}{2} U\). (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius \(r\) around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches from afar with kinetic energy \(T_{2} .\) When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius \(r^{\prime} .\) (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy \(E\) long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables \(T_{2}, r,\) and \(r^{\prime}\).

A charge \(q\) in a uniform electric field \(\mathbf{E}_{0}\) experiences a constant force \(\mathbf{F}=q \mathbf{E}_{0}\). (a) Show that this force is conservative and verify that the potential energy of the charge at position \(\mathbf{r}\) is \(U(\mathbf{r})=-q \mathbf{E}_{0} \cdot \mathbf{r}\). (b) By doing the necessary derivatives, check that \(\mathbf{F}=-\nabla U\).

Both the Coulomb and gravitational forces lead to potential energies of the form \(U=\gamma / | \mathbf{r}_{1}-\) \(\mathbf{r}_{2} |,\) where \(\gamma\) denotes \(k q_{1} q_{2}\) in the case of the Coulomb force and \(-G m_{1} m_{2}\) for gravity, and \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) are the positions of the two particles. Show in detail that \(-\nabla_{1} U\) is the force on particle 1 and \(-\nabla_{2} U\) that on particle 2.

In one dimension, it is obvious that a force obeying Hooke's law is conservative (since \(F=-k x\) depends only on the position \(x,\) and this is sufficient to guarantee that \(F\) is conservative in one dimension). Consider instead a spring that obeys Hooke's law and has one end fixed at the origin, but whose other end is free to move in all three dimensions. (The spring could be fastened to a point in the ceiling and be supporting a bouncing mass \(m\) at its other end, for instance.) Write down the force \(\mathbf{F}(\mathbf{r})\) exerted by the spring in terms of its length \(r\) and its equilibrium length \(r_{\mathrm{o}} .\) Prove that this force is conservative. [Hints: Is the force central? Assume that the spring does not bend.]

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