Stopping a Rotating System with Tangential Force: A Simple Calculation

This article explains a straightforward problem involving the application of a tangential force to a rotating system, where the objective is to determine the time it takes to stop the motion. We will analyze the problem using Newton's second law for rotational motion, focusing on the principles of angular momentum and torque.

Introduction

The problem at hand involves a circle with a center bearing that supports twelve 100-gram masses evenly arrayed around it. The bearing has an arc speed of 1 m/sec. The challenge posed is to find out how long it would take for a 1 N force applied tangentially to stop the motion.

Calculation of Initial Conditions

The first step is to determine the combined mass and the initial momentum of the system. Given that each mass is 100 grams (0.1 kg), the total mass of the twelve masses combined is 1.2 kg. Assuming a momentum of 1.2 kg·m/s, the time ( t ) to bring the system to rest can be calculated as follows:

Combined Mass (M): 1.2 kg

Momentum (P): 1.2 kg·m/s 1.2 kg·m/s

Time (t): ( t frac{1.2 text{ kg·m/s}}{1.2 text{ kg}} 1.2 text{ seconds} )

Application of Newton's Second Law for Rotational Motion

To apply Newton's second law for rotational motion, we need to consider the following relationships:

Equation of Torque (T)

First, we have:

T Ialpha I frac{Delta omega}{Delta t}

Or alternatively:

Tt Iomega

Substituting omega frac{v}{R} and T FR into the equation:

FRt 12mR^2 frac{v}{R}

This simplifies to:

Ft 12mv

Substituting the given values (F 1 N, m 1.2 kg, v 1 m/s):

1 Nt 1.2 kg cdot 1 m/s

Therefore, t 1.2 text{ seconds}

Simple Linear Mechanics Perspective

The problem of rotating masses can be directly related to linear mechanics, where the force is applied tangentially and thus has the same direction as the tangential velocity. The circular motion is only for context; it does not affect the calculation of stopping time.

Considering the motion in a straight line, the principles of linear mechanics apply, and the time to stop the motion remains the same as the calculated value from the rotational motion perspective.

Further Considerations

Note that if the rotating mass is other than a rim (such as a solid disk), the calculation of time to stop the motion would not be the same. The moment of inertia ((I 12mR^2)) for a different shape would change, leading to a different time required to stop the motion.

Therefore, for a rim (circular mass distribution), the time to stop is indeed 1.2 seconds, but for other shapes, the time would be different due to a different moment of inertia.