Problem 80 - Physics for Scientists and Engineers with Modern Physics

Question

A flat puck (mass \(M\) ) is revolved in a circle on a frictionless air hockey table top, and is held in this orbit by a light cord which is connected to a dangling mass (mass \(m\) ) through a central hole as shown in Fig. \(5-48 .\) Show that the speed of the puck is given by \(v=\sqrt{m g R / M}\)

Step-by-Step Solution

Answer

The speed of the puck is \(v = \sqrt{\frac{mgR}{M}}\).

Step 1: Identify Forces at Play

The system consists of two masses: a flat puck of mass \(M\) on a frictionless table, and a dangling mass \(m\) hanging off the table. The two are connected by a cord through a central hole. The force keeping the puck moving in a circle is the tension \(T\) in the cord, caused by the weight \(mg\) of the dangling mass.

Step 2: Relate Tension and Weight

Since the system is in equilibrium, the tension \(T\) in the cord must be equal to the gravitational force on the dangling mass: \(T = mg\).

Step 3: Apply Newton's Second Law for Circular Motion

For the puck in circular motion, the tension \(T\) is what provides the centripetal force needed to keep it moving in a circle with radius \(R\). According to Newton's Second Law, this centripetal force is given by \(F_c = \frac{M v^2}{R}\), where \(v\) is the speed of the puck.

Step 4: Equate Forces

Since the tension provides the centripetal force, set the expressions for tension equal to the centripetal force expression: \(mg = \frac{M v^2}{R}\).

Step 5: Solve for Speed of the Puck

Rearrange the equation \(mg = \frac{M v^2}{R}\) to solve for \(v\). Multiply both sides by \(R\) to get \(mgR = Mv^2\). Then divide both sides by \(M\) to isolate \(v^2\): \(v^2 = \frac{mgR}{M}\). Finally, take the square root of both sides to find \(v\): \(v = \sqrt{\frac{mgR}{M}}\).

Key Concepts

Centripetal Force

In circular motion, an object continuously changes direction to follow a circular path. To maintain this motion, a force that acts towards the center of the circle is required. This force is known as the "centripetal force." For the puck in our scenario, the tension in the cord serves as the centripetal force. It constantly pulls the puck towards the center, keeping it on its circular path.

The equation for centripetal force is given by:

• \( F_c = \frac{M v^2}{R} \)

Where:

• \( M \) is the mass of the object in motion (the puck),
• \( v \) is the speed of the object, and
• \( R \) is the radius of the circle.

For circular motion, the centripetal force does not perform work on the object since it acts perpendicular to the velocity. Its role is purely to change the direction of the object's motion, not its speed.

Newton's Second Law

Newton's Second Law of Motion is a fundamental principle that illustrates the relationship between an object's mass, its acceleration, and the applied force. It is succinctly expressed by the equation:

• \( F = ma \)

Where:

• \( F \) is the net force acting on the object,
• \( m \) is the object's mass, and
• \( a \) is the acceleration of the object.

In our exercise, Newton’s Second Law applies to the centripetal motion of the puck. Here, the acceleration is centripetal, pointing towards the circle's center. Thus, the net force acting as the centripetal force (tension in the cord) is given by \( T = \frac{M v^2}{R} \).

This relationship further aids in understanding how the tension caused by the dangling mass is crucial for determining the speed of the puck, as both forces work together to satisfy Newton's law in a dynamic situation.

Equilibrium in Physics

Equilibrium in physics refers to the state where all forces acting on a system are balanced, resulting in no net force. In the context of our exercise, the equilibrium is between the gravitational force acting on the hanging mass and the tension in the cord. For these forces to be in equilibrium:

• The tension must exactly match the gravitational force due to the hanging mass.

Mathematically, this is represented as:

• \( T = mg \)

Where:

• \( T \) is the tension in the cord,
• \( m \) is the mass of the hanging object, and
• \( g \) is the acceleration due to gravity.

When both forces are equal, the system does not accelerate in the vertical direction because the forces cancel each other out. This scenario ensures that the puck's tension is consistently supplied to act as the needed centripetal force, allowing the horizontal circular motion to proceed with constant velocity. Understanding equilibrium is critical since it provides the basis for analyzing and solving the dynamics involved in the system.