Problem 18 - An Introduction to Mechanics

Question

Sand runs from a hopper at constant rate \(d m / d t\) onto a horizontal conveyor belt driven at constant speed \(V\) by a motor. ( \(a\) ) Find the power needed to drive the belt. (b) Compare the answer to \((a)\) with the rate of change of kinetic energy of the sand. Can you account for the difference?

Step-by-Step Solution

Answer

The power needed is \( \left( \dfrac{dm}{dt} \right) V^2 \), twice the sand's kinetic energy rate change.

Step 1: Understand the Problem

In this problem, sand falls at a constant rate onto a conveyor belt moving at constant speed. Our task is to calculate the power needed to drive the conveyor belt and to compare it with the rate of change of kinetic energy of the sand.

Step 2: Calculate the Power Needed to Drive the Belt

Power is the rate of work done. When sand falls on the conveyor belt, it gains velocity and the belt does work on it. The force required to accelerate the sand to speed \(V\) is \(F = d \left( \dfrac{m}{dt} \right) V\). Therefore, the power can be expressed as:\[P = F \cdot V = \left( \dfrac{dm}{dt} \right) V^2.\]

Step 3: Determine the Rate of Change of Kinetic Energy of the Sand

The kinetic energy gained by the sand as it gains speed \(V\) from rest is given by:\[KE = \dfrac{1}{2} m V^2.\]Thus, the rate of change of kinetic energy with respect to time is:\[\dfrac{d(KE)}{dt} = \dfrac{1}{2} \left( \dfrac{dm}{dt} \right) V^2.\]

Step 4: Compare and Analyze

The power required to drive the belt is \( \left( \dfrac{dm}{dt} \right) V^2 \) and the rate of change of kinetic energy of the sand is \( \dfrac{1}{2} \left( \dfrac{dm}{dt} \right) V^2 \). The power needed is twice the rate of change of kinetic energy because additional work is needed to sustain the movement of the belt's own mass and overcome any friction in the system.

Key Concepts

Power

Power is an essential concept in mechanics that refers to the rate at which work is done, or energy is transferred, over time. In the context of the conveyor belt problem, power is needed to keep the belt moving against forces such as friction and to accelerate the sand as it falls onto the belt.
When calculating the power required to drive a conveyor belt, we consider the force that must be exerted on the sand to bring it up to the belt's speed. This force is calculated using the expression \( F = \left( \dfrac{dm}{dt} \right) V \), where \( \dfrac{dm}{dt} \) is the rate at which mass (sand) falls onto the belt, and \( V \) is the constant speed of the belt.
The power \( P \) is the product of this force and the velocity, given by \( P = F \cdot V = \left( \dfrac{dm}{dt} \right) V^2 \). This shows that power is heavily dependent on both the mass flow rate and the speed of the conveyor belt.

• Power is measured in watts (W), where 1 watt is equivalent to 1 joule per second.
• In this scenario, our calculated power represents the energy required per second to both maintain the movement of the belt and impart kinetic energy to the sand.

Kinetic Energy

Kinetic energy relates to the energy an object possesses due to its motion. In this exercise, the sand particles acquire kinetic energy as they are accelerated to match the speed of the conveyor belt.
The formula for kinetic energy of an object with mass \( m \) moving at velocity \( v \) is \( KE = \dfrac{1}{2} m v^2 \). As the sand grains fall onto the conveyor belt from rest, they develop speed \( V \), resulting in a kinetic energy gain described by this formula.
We look at how the kinetic energy changes with time as more sand accumulates on the belt, which can be calculated as \( \dfrac{d(KE)}{dt} = \dfrac{1}{2} \left( \dfrac{dm}{dt} \right) V^2 \).
This rate of change is only half of the power needed to drive the belt, showing that not all the energy from the motor contributes to the sand's kinetic energy.

• Kinetic energy is a scalar quantity measured in joules (J).
• This part of the energy calculation only considers the energy needed to bring the sand up to speed, separate from what is needed to overcome friction and to move the belt itself.

Conveyor Belt

Conveyor belts are widely used mechanical systems that transport materials from one place to another. They are found in various settings, such as warehouses, manufacturing plants, and even in the food industry.
The belt itself is a loop of material that moves continuously over rollers or pulleys, driven by a motor. In our scenario, the belt speed \( V \) is constant, simplifying calculations for power and kinetic energy required by the system.
Analyzing the conveyor belt's power requirements in this context requires considering both the kinetic energy imparted to materials and any resistance forces like friction or drag.One of the significant insights from this problem is understanding the interplay between the belt's motion and the materials it transports.

• The physics principles at play ensure the conveyor belt achieves a smooth transport of material while dealing with mechanical resistance.
• The energy balance calculated in this case exemplifies how mechanical systems require additional power to maintain steady operations beyond just the energy change in the material being moved.