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A meter stick is found to balance at the \(49.7-\mathrm{cm}\) mark when placed on a fulcrum. When a \(50.0\)-gram mass is attached at the \(10.0-\mathrm{cm}\) mark, the fulcrum must be moved to the \(39.2-\mathrm{cm}\) mark for balance. What is the mass of the meter stick?

Short Answer

Expert verified
The mass of the meter stick is \(100.0 \, \mathrm{g}\).

Step by step solution

01

Identify variables

We have the following information: the initial balance point without any load is \(49.7 \, \mathrm{cm}\), the mass added is \(50.0 \, \mathrm{g}\) at the \(10.0 \, \mathrm{cm}\) mark, and the new balance point when the \(50.0 \, \mathrm{g}\) mass is added is \(39.2 \, \mathrm{cm}\). We need to find the mass of the meter stick.
02

Apply principle of moments

The meter stick and the \(50.0 \, \mathrm{g}\) mass are balanced, which means the sum of moments about the fulcrum is zero. This gives us the equation:\[(meter \, stick \, mass) \times (distance \, to \, fulcrum) = (added \, mass) \times (distance \, to \, fulcrum)\]or\[(M \times 39.2) = (50 \times (50-10)) \]where M is the mass of the meter stick in grams.
03

Solve for the meter stick mass

To find the mass of the meter stick, we rearrange the equation above:\[M = \frac{50 \times (50-10)}{39.2}\]And upon calculation, we find that M = \(100.0 \, g\). The mass of the meter stick is \(100.0 \, \mathrm{g}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Principle of Moments
The principle of moments, also known as Varignon's theorem, is a foundational concept in physics that applies to conditions of equilibrium. According to this principle, for an object to be in equilibrium, the sum of the clockwise moments about any pivot point must be equal to the sum of the counterclockwise moments around that same pivot.
For example, if you hold a meter stick horizontally on your finger, the moment (which is the product of the force applied and the distance from the pivot) created by the weight of the stick to one side of your finger must be equal to the moment on the other side. This principle helps us understand how objects balance and is vital when solving problems involving levers and seesaws.

Application to the Meter Stick Problem

In the problem presented, we first identified the pivot points (fulcrums) and then applied the principle of moments to find that the meter stick's weight times the distance from the new fulcrum must equal the 50.0-gram mass times its distance from the fulcrum. This balance of moments allowed us to solve for the unknown mass of the meter stick.
Static Equilibrium
Static equilibrium refers to a state where an object is at rest and remains in that state unless acted upon by an external force. For an object to be in static equilibrium, both the net force and the net torque (moment of force) acting on it must be zero.
In the context of balancing objects such as our meter stick, static equilibrium is achieved when the meter stick does not rotate or move because the torques acting on it are in balance. This is exactly what you observe when the meter stick balances on the fulcrum without tipping over.

Understanding with the Meter Stick

For the meter stick balance problem, static equilibrium was reached twice: once without the additional mass and a second time with the 50.0-gram mass, when the fulcrum was adjusted. This demonstrates that despite changing conditions, static equilibrium can be reestablished by altering the position of the fulcrum.
Torque Balance
Torque balance is a crucial aspect of rotational mechanics, often used alongside the concepts of static equilibrium and the principle of moments. Torque, also known as moment of force, is the rotational equivalent of linear force and is calculated by multiplying the force applied by the distance from the pivot or fulcrum at which it acts.
To achieve a torque balance, the sum of all torques acting in one direction around a pivot must be equal to the sum of torques acting in the opposite direction. This is essential for an object to be in rotational equilibrium, meaning it doesn't spin uncontrollably.

Relevance to Our Exercise

In the solved exercise, torque balance is demonstrated when the meter stick is in static equilibrium, both before and after the 50.0-gram mass is added. By balancing the torques around the fulcrum, we were able to deduce the mass of the meter stick based on the given distances and the mass of the added weight.
Center of Mass
The center of mass is the point at which the mass of a body is considered to be concentrated for the purposes of analysis under external forces and moments. In simple terms, it's the 'average' location of all the mass in an object.
For uniform objects, like a plain meter stick with no additional weights, the center of mass is usually at the geometric center. However, when weights are added or the distribution of mass changes, the center of mass shifts accordingly. This is important for predicting the behavior of the object when supported or suspended.

Application in the Problem

In the scenario with the meter stick, the center of mass initially lies at the 49.7 cm mark. When the 50.0-gram weight is added to the 10.0 cm mark, the balance point shifts, indicating that the center of mass has moved. By understanding this concept and its implications on balance, we can make precise calculations as seen in the steps outlined in the exercise. Understanding the center of mass provides insight into how to achieve balance by aligning it with the pivot point.

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Most popular questions from this chapter

QIC S A cylinder with moment of inertia \(I_{1}\) rotates with angular velocity \(\omega_{0}\) about a frictionless vertical axle. A second cylinder, with moment of inertia \(I_{2}\), initially not rotating, drops onto the first cylinder (Fig. P8.65). Because the surfaces are rough, the two cylinders eventually reach the same angular speed \(\omega\). (a) Calculate \(\omega\). (b) Show that kinetic energy is lost in this situation, and calculate the ratio of the final to the initial kinetic energy.

Halley's comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being \(0.59 \mathrm{~A} . \mathrm{U} .\) and its greatest distance being \(35 \mathrm{~A} . \mathrm{U}\). ( \(1 \mathrm{~A} . \mathrm{U}\). is the Earth-Sun distance). If the comet's speed at closest approach is \(54 \mathrm{~km} / \mathrm{s}\), what is its speed when it is farthest from the Sun? You may neglect any change in the comet's mass and assume that its angular momentum about the Sun is conserved.

A space station shaped like a giant wheel has a radius of \(100 \mathrm{~m}\) and a moment of inertia of \(5.00 \times 10^{8} \mathrm{~kg} \cdot \mathrm{m}^{2}\) A crew of 150 lives on the rim, and the station is rotating so that the crew experiences an apparent acceleration of \(1 g\) (Fig. P8.64).When 100 people move to the center of the station for a union meeting, the angular speed changes. What apparent acceleration is experienced by the managers remaining at the rim? Assume the average mass of a crew member is \(65.0 \mathrm{~kg}\).

A large grinding wheel in the shape of a solid cylinder of radius \(0.330 \mathrm{~m}\) is free to rotate on a frictionless, vertical axle. A constant tangential force of \(250 \mathrm{~N}\) applied to its edge causes the wheel to have an angular acceleration of \(0.940 \mathrm{rad} / \mathrm{s}^{2}\). (a) What is the moment of inertia of the wheel? (b) What is the mass of the wheel? (c) If the wheel starts from rest, what is its angular velocity after \(5.00 \mathrm{~s}\) have elapsed, assuming the force is acting during that time?

S This is a symbolic version of problem 72. Two astronauts (Fig. P8.72), each having a mass \(M\), are connected by a rope of length \(d\) having negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed \(v\). (a) Calculate the magnitude of the angular momentum of the system by treating the astronauts as particles. (b) Calculate the rotational energy of the system. By pulling on the rope, the astronauts shorten the distance between them to \(d / 2\). (c) What is the new angular momentum of the system? (d) What are their new speeds? (e) What is the new rotational energy of the system? (f) How much work is done by the astronauts in shortening the rope?

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