Question

A hockey player hits a puck of mass \(m_{P}\) into a semicircular groove in the ice of radius \(R\). The ice has coefficient of Coulomb friction \(\mu_{c}\). What is the impulse that must be imparted to the puck at the entrance to the groove such that it makes it around the semicircle exactly once before stopping?

Short answer

The impulse required is \( J = m_{P} \cdot \sqrt{2 \mu_{c} g \pi R} \).

Step-by-step solution

Understanding the Problem

We need to calculate the impulse required to make the puck move around a semicircular groove with the given properties. The impulse will compensate for velocity loss due to friction.

Determine Frictional Force

The frictional force \( F_{f} \) acting on the puck as it moves along the groove can be calculated using \( F_{f} = \mu_{c} \cdot m_{P} \cdot g \), where \( g \) is the acceleration due to gravity.

Calculate Work Done by Friction

The work done by the frictional force as the puck travels the semicircular path of radius \( R \) is given by \( W = F_{f} \cdot \pi R \), since the path length of the semicircle is \( \pi R \).

Relate Work to Initial Kinetic Energy

To make the puck stop exactly after moving around the semicircle, the work done by friction will be equal to the initial kinetic energy imparted by the impulse. Set \( W = \frac{1}{2} m_{P} v^2 \), where \( v \) is the initial velocity due to the impulse.

Solve for Initial Velocity

From \( \frac{1}{2} m_{P} v^2 = \mu_{c} \cdot m_{P} \cdot g \cdot \pi R \), we can solve for \( v \) to get the initial velocity \( v = \sqrt{2 \mu_{c} g \pi R} \).

Calculate Impulse

Impulse \( J \) is the change in momentum, defined as \( J = m_{P} \cdot v \). Substituting the value of \( v \) from the previous step, we get \( J = m_{P} \cdot \sqrt{2 \mu_{c} g \pi R} \).

Key concepts

Impulse

Impulse, in physics, is defined as the change in momentum of an object when a force is applied over a period of time. In this exercise, the impulse is the push given to the hockey puck to ensure it makes it around the semicircle. This impulse allows the puck to overcome frictional forces and gain initial kinetic energy necessary to move along the groove.

The formula for impulse is given by:

• Impulse ( \( J \)) = Change in momentum = Final momentum - Initial momentum
• If an initial velocity ( \( v \)) is imparted to the stationary puck of mass ( \( m_{P} \)), then impulse can also be calculated as:
• \( J = m_{P} \, v \)

This is what we used to determine how much push was needed to inject enough energy into the puck so it could travel completely around the track before coming to rest.

Frictional Force

Frictional force arises due to the surfaces in contact when the puck slides along the ice. In our scenario, the frictional force is the force that resists the puck's motion through the groove. It is due to the roughness of the ice and is proportional to the normal force, which in this case is the product of the puck's mass and gravity.

The formula to calculate frictional force (\( F_{f} \)) is:

• \( F_{f} = \mu_{c} \cdot m_{P} \cdot g \)

Here, \( \mu_{c} \) is the coefficient of Coulomb friction, \( m_{P} \) is the mass of the puck, and \( g \) is the acceleration due to gravity (approximately \( 9.81\, \text{m/s}^2 \)).
This force is always directed opposite to the direction of motion, reducing the puck's kinetic energy over time.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. When the impulse is applied to the puck, it converts that impulse into kinetic energy, propelling it around the groove. The kinetic energy must be enough to counteract the work done by friction over the path of the semicircle.

The formula for kinetic energy (\( KE \)) of an object is given by:

• \( KE = \frac{1}{2} m_{P} v^2 \)

Here, \( m_{P} \) is the mass of the puck and \( v \) is its velocity. This energy gets entirely consumed by the frictional work done as the puck traverses the semicircle. In our solution, equating the work done by friction to the puck's initial kinetic energy allowed us to calculate the required velocity.

Coulomb Friction

Coulomb friction, named after Charles-Augustin de Coulomb, is a simple model that defines friction as proportional to the normal force. It describes a linear relationship where the frictional force remains constant regardless of speed, as long as the surfaces remain unchanged. In this exercise, Coulomb friction helps us understand the resistance the puck faces while moving along the ice surface.

Mathematically, Coulomb friction is represented as:

• \( F_{f} = \mu_{c} \cdot N = \mu_{c} \cdot m_{P} \cdot g \)

Where \( \mu_{c} \) is the coefficient of friction, and \( N \) is the normal force, which is the product of mass \( m_{P} \) and the acceleration due to gravity \( g \).
Understanding this principle is vital for calculating the work done by friction and thus the initial impulse required.