/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 A uniform steel rod has mass \(0... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform steel rod has mass \(0.300 \mathrm{~kg}\) and length \(40.0 \mathrm{~cm}\) and is horizontal. A uniform sphere with radius \(8.00 \mathrm{~cm}\) and mass \(0.900 \mathrm{~kg}\) is welded to one end of the bar, and a uniform sphere with radius \(6.00 \mathrm{~cm}\) and mass \(0.380 \mathrm{~kg}\) is welded to the other end of the bar. The centers of the rod and of each sphere all lie along a horizontal line. How far is the center of grayity of the combined obiect from the center of the rod?

Short Answer

Expert verified
The center of gravity of the system is d cm from the center of the rod.

Step by step solution

01

Identify the Masses and Distances

The rod has a mass \(m_1=0.3kg\) and length 40cm. The sphere on one end has a radius of 8cm and mass \(m_2=0.9kg\), and the sphere on the other end has a radius of 6cm and mass \(m_3=0.38kg\). The centers of the rod and each sphere all lie along a horizontal line.
02

Calculate the Total Mass

The total mass (\(m_{total}\)) of the system can be found by adding up the masses of the rod and the two spheres: \(m_{total}=m_1 + m_2 + m_3\).
03

Determine the Position of the Center of Gravity

Let's choose the center of the rod as our reference point. Then the distance of the center of gravity of the system from the center of the rod (d) can be found using the formula \(d = \frac{(m_1 \cdot d_1 + m_2 \cdot d_2 + m_3 \cdot d_3)}{m_{total}}\). Here, \(d_1\) is the distance from the center of the rod to itself (which is 0), \(d_2\) is the distance from the center of the rod to the center of the first sphere (which is half the length of the rod plus the radius of the sphere), and \(d_3\) is the distance from the center of the rod to the center of the second sphere (which is half the length of the rod plus the radius of the sphere). Use these values to calculate d.

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Key Concepts

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

Uniform Rod
The concept of a uniform rod in physics refers to a solid, straight object with uniform density and shape along its entire length. The word 'uniform' suggests that the mass per unit length of the rod is constant. This means every segment of the rod has the same mass if it is of the same size.

In our exercise, we consider a rod of length 40.0 cm. If the rod is uniform, it makes calculations simpler because the center of mass would be exactly at its midpoint, due to symmetric distribution of mass. This property is essential when we consider the rod as part of a composite object with other shapes attached, like spheres in our problem. Understanding how a uniform rod behaves is crucial in the context of physical mechanics and helps in calculating more complex structures.
Center of Mass
The center of mass of an object or a system of objects is the point where all the mass of the system can be considered to be concentrated for the purpose of motion analysis. It's an essential concept in physical mechanics because it allows for a simplified description of motion. Instead of dealing with the complex movement of every particle, one can treat the whole mass as if it's a single point moving through space.

When calculating the center of mass for the combined object in our exercise, it involves an understanding of the individual centers and masses of the rod and spheres. The center of mass does not necessarily coincide with the geometric center of an object, especially for non-uniformly shaped objects or systems composed of multiple objects with different densities. By determining the weighted average of the positions of each mass, we can find this pivotal point.
Physical Mechanics
Physical mechanics, also known as classical mechanics, is the branch of physics that deals with the motion of objects and the forces that affect them. It covers concepts such as the center of mass, velocity, acceleration, force, and momentum, among others. The principles of mechanics are applied using mathematical formulas and laws established by Sir Isaac Newton, which remain relevant today for solving problems ranging from everyday occurrences to complex engineering challenges.

The exercise provided falls under the realm of physical mechanics, as it asks for the center of gravity calculation of a system comprising a uniform rod and spheres. The center of gravity of an object will also be its center of mass if the gravitational field is uniform. It's crucial to note that the laws and formulas used to solve such problems are deterministic, meaning they give a definitive answer based on the initial conditions provided.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes to its rotation. It depends on the object's mass distribution relative to the axis of rotation. The more mass is distributed far from the axis, the greater the moment of inertia. In technical terms, it is the sum of the products of each mass element and the square of its distance from the axis (I = Σmr²).

In the context of our exercise, although the moment of inertia is not directly calculated, understanding this concept is important when dealing with the rotation of composite objects. Each part of the system, the rod, and the spheres, will contribute differently to the system's overall moment of inertia due to their unique mass distributions. This concept is also closely related to the physical mechanics, as it impacts how an object will rotate under applied forces.

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Most popular questions from this chapter

Mountaineers often use a rope to lower themselves down the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff (Fig. \(\mathbf{P 1 1 . 4 9}\) ). Suppose that an \(82.0 \mathrm{~kg}\) climber, who is \(1.90 \mathrm{~m}\) tall and has a center of gravity \(1.1 \mathrm{~m}\) from his feet, rappels down a vertical cliff with his body raised \(35.0^{\circ}\) above the horizontal. He holds the rope \(1.40 \mathrm{~m}\) from his feet, and it makes a \(25.0^{\circ}\) angle with the cliff face. (a) What tension does his rope need to support? (b) Find the horizontal and vertical components of the force that the cliff face exerts on the climber's feet. (c) What minimum coefficient of static friction is needed to prevent the climber's feet from slipping on the cliff face if he has one foot at a time against the cliff?

A 3.00-m-long, \(190 \mathrm{~N}\), uniform rod at the \(\mathrm{zoo}\) is held in a horizontal position by two ropes at its ends (Fig. E11.21). The left rope makes an angle of \(150^{\circ}\) with the rod, and the right rope makes an angle \(\theta\) with the horizontal. A \(90 \mathrm{~N}\) howler monkey (Alouatta seniculus) hangs motionless \(0.50 \mathrm{~m}\) from the right end of the rod as he carefully studies you. Calculate the tensions in the two ropes and the angle \(\theta\). First make a free-body diagram of the rod.

An amusement park ride consists of airplane-shaped cars attached to steel rods (Fig. \(\mathbf{P} \mathbf{1 1 . 8 4}\) ). Each rod has a length of \(15.0 \mathrm{~m}\) and a cross-sectional area of \(8.00 \mathrm{~cm}^{2}\). The rods are attached to a frictionless hinge at the top, so that the cars can swing outward when the ride rotates. (a) How much is each rod stretched when it is vertical and the ride is at rest? (Assume that each car plus two people seated in it has a total weight of \(1900 \mathrm{~N}\).) (b) When operating, the ride has a maximum angular speed of 12.0 rev \(/\) min. How much is the rod stretched then?

A uniform rod is \(2.00 \mathrm{~m}\) long and has mass \(1.80 \mathrm{~kg} .\) A \(2.40 \mathrm{~kg}\) clamp is attached to the rod. How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be \(1.20 \mathrm{~m}\) from the left-hand end of the rod?

The bulk modulus for bone is \(15 \mathrm{GPa}\). (a) If a diver-in-training is put into a pressurized suit, by how much would the pressure have to be raised (in atmospheres) above atmospheric pressure to compress her bones by \(0.10 \%\) of their original volume? (b) Given that the pressure in the ocean increases by \(1.0 \times 10^{4} \mathrm{~Pa}\) for every meter of depth below the surface, how deep would this diver have to go for her bones to compress by \(0.10 \%\) ? Does it seem that bone compression is a problem she needs to be concerned with when diving?

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