Question

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

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

Step-by-step solution

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 \(h\):\[v = \sqrt{2gh}\]

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 \(N\):\[N = \frac{m(2gh)}{R} - mg\cos(\theta)\]

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:\[h = R(1 - \frac{2}{3}) = \frac{R}{3}\]

Conclusion

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

Key concepts

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.