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Chapter 11: Problem 5

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 = 危mr虏).

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

The compressive strength of our bones is important in everyday life. Young's modulus for bone is about \(1.4 \times 10^{10} \mathrm{~Pa}\). Bone can take only about a \(1.0 \%\) change in its length before fracturing. (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is \(3.0 \mathrm{~cm}^{2} ?\) (This is approximately the cross-sectional area of a tibia, or shin bone, at its narrowest point.) (b) Estimate the maximum height from which a \(70 \mathrm{~kg}\) man could jump and not fracture his tibia. Take the time between when he first touches the floor and when he has stopped to be \(0.030 \mathrm{~s}\), and assume that the stress on his two legs is distributed equally.

You are to use a long, thin wire to build a pendulum in a science mu- seum. The wire has an unstretched length of \(22.0 \mathrm{~m}\) and a cir- cular cross section of \(\begin{array}{lll}\text { diameter } & 0.860 & \mathrm{~mm}\end{array}\) it is made of an alloy that has a large breaking stress. One end of the wire will be attached to the ceiling, and a \(9.50 \mathrm{~kg}\) metal sphere will be attached to the other end. As the pendulum swings back and forth, the wire's maximum angular displacement from the vertical will be \(36.0^{\circ}\). You must determine the maximum amount the wire will stretch during this motion. So, before you attach the metal sphere, you suspend a test mass (mass \(m\) ) from the wire's lower end. You then measure the increase in length \(\Delta l\) of the wire for several different test masses. Figure \(\mathbf{P} 1 \mathbf{1} .86,\) a graph of \(\Delta l\) versus \(m\) shows the results and the straight line that gives the best fit to the data. The equation for this line is \(\Delta l=(0.422 \mathrm{~mm} / \mathrm{kg}) m\) (a) Assume that \(g=9.80 \mathrm{~m} / \mathrm{s}^{2},\) and use Fig. \(\mathrm{P} 11.86\) to calculate Young's modulus \(Y\) for this wire. (b) You remove the test masses, attach the \(9.50 \mathrm{~kg}\) sphere, and release the sphere from rest, with the wire displaced by \(36.0^{\circ} .\) Calculate the amount the wire will stretch as it swings through the vertical. Ignore air resistance.

In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of \(90.8 \mathrm{~N}\) is applied perpendicular to each end. If the diameter of the wire is \(1.84 \mathrm{~mm},\) what is the breaking stress of the alloy?

An engineer is designing a conveyor system for loading hay bales into a wagon (Fig. \(\mathbf{P} 1 \mathbf{1} .79)\). Each bale is \(0.25 \mathrm{~m}\) wide, \(0.50 \mathrm{~m}\) high, and \(0.80 \mathrm{~m}\) long (the dimension perpendicular to the plane of the figure), with mass \(30.0 \mathrm{~kg}\). The center of gravity of each bale is at its geometrical center. The coefficient of static friction between a bale and the conveyor belt is \(0.60,\) and the belt moves with constant speed. (a) The angle \(\beta\) of the conveyor is slowly increased. At some critical angle a bale will tip (if it doesn't slip first), and at some different critical angle it will slip (if it doesn't tip first). Find the two critical angles and determine which happens at the smaller angle. (b) Would the outcome of part (a) be different if the coefficient of friction were \(0.40 ?\)

A therapist tells a \(74 \mathrm{~kg}\) patient with a broken leg that he must have his leg in a cast suspended horizontally. For minimum discomfort, the leg should be supported by a vertical strap attached at the center of mass of the leg-cast system (Fig. \(\mathbf{P} 11.55\) ). To comply with these instructions, the patient consults a table of typical mass distributions and finds that both upper legs (thighs) together typically account for \(21.5 \%\) of body weight and the center of mass of each thigh is \(18.0 \mathrm{~cm}\) from the hip joint. The patient also reads that the two lower legs (including the feet) are \(14.0 \%\) of body weight, with a center of mass \(69.0 \mathrm{~cm}\) from the hip joint. The cast has a mass of \(5.50 \mathrm{~kg}\), and its center of mass is \(78.0 \mathrm{~cm}\) from the hip joint. How far from the hip joint should the supporting strap be attached to the cast?
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