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A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

Short Answer

Expert verified
(a) \(\frac{ML^2}{12}\); (b) \(\frac{7ML^2}{12}\).

Step by step solution

01

Analyze the Rod Configuration

The rod is bent at its center into two segments, each of length \(\frac{L}{2}\), and these segments are perpendicular to each other forming an 'L' shape. The mass of each segment is \(\frac{M}{2}\) given the uniform distribution of mass.
02

Moment of Inertia at the Corner

Consider the point where the two segments meet, creating a single axis perpendicular to the plane of the rod. Treat each segment separately and use the formula for the moment of inertia of a rod about its end point, \(I = \frac{1}{3}mL^2\). Here, \(m = \frac{M}{2}\) and \(L = \frac{L}{2}\). Add up the inertia contributions from both segments:\[I = 2 \times \left(\frac{1}{3} \times \frac{M}{2} \times \left(\frac{L}{2}\right)^2\right) = \frac{1}{3} \frac{ML^2}{4} = \frac{ML^2}{12}.\] Therefore, \(I_{corner} = \frac{ML^2}{12}.\)
03

Moment of Inertia at the Midpoint of the Ends

Locate the midpoint of the line connecting the outer ends of the 'L' shape. The segments being perpendicular form a right-angled isosceles triangle with the diagonal from the origin of the corner to the center where the inertia is being considered. Thus, use the parallel axis theorem:\[I_{midpoint} = I_{corner} + M \times R^2,\] where \(R = \frac{L}{\sqrt{2}}.\) Therefore,\[I_{midpoint} = \frac{ML^2}{12} + M \left(\frac{L}{\sqrt{2}}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{2} = \frac{7ML^2}{12}.\]

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Key Concepts

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

Parallel Axis Theorem
The parallel axis theorem is a critical tool in calculating moments of inertia of objects around axes not passing through the center of mass. It helps us find the moment of inertia about any axis parallel to an axis through the center of mass. The formula is given by:\[ I = I_{cm} + Md^2 \] where:
  • \( I \) is the moment of inertia about the new axis.
  • \( I_{cm} \) is the moment of inertia of the object around the center of mass axis.
  • \( M \) is the total mass of the object.
  • \( d \) is the perpendicular distance between the two axes.
In our problem, when considering the midpoint between the ends of the bent rod, the parallel axis theorem is applied. We first calculate the moment of inertia at the corner where the two segments meet and then account for the diagonal distance to the new axis. Understanding this shift in axis is key in many practical applications, such as engineering and physics problems.
Uniform Rod
A uniform rod refers to a rod with mass distributed evenly along its length. This characteristic simplifies calculations of physical properties, such as moments of inertia. For calculating moments of inertia of a uniform rod about different axes, we use various standard formulas.In this situation, the rod is bent into an 'L' shape, where each segment's mass and length are halved due to the uniform distribution. This condition is essential in calculating the individual contributions of inertia from each segment about a specific axis.For a straight uniform rod with length \( L \) and mass \( M \), the moment of inertia about an axis through one end perpendicular to its length is given by:\[ I = \frac{1}{3} ML^2 \]This formula is adapted for our case, where both segments are \( \frac{L}{2} \) in length and \( \frac{M}{2} \) in mass.
Perpendicular Axis
The configuration in our exercise requires considering two segments of a rod arranged perpendicularly. Each segment contributes individually to the overall moment of inertia about an axis that is also perpendicular to the plane formed by the segments. The perpendicular setup means we can't just treat the rod as one linear piece. Instead, we handle each segment separately. In calculation, this configuration allows us to treat moments about each segment's end and then consider them collectively. This setup is the basis of the concept called the "perpendicular axis theorem" which applies mainly in scenarios involving symmetry across two dimensions, which then integrates with the parallel axis theorem for 3D calculations.
Rotational Dynamics
Rotational dynamics explores the motion of bodies that rotate about an axis, providing insights into aspects like angular momentum, torque, and moment of inertia. The current exercise showcases how a modified shape of a uniform rod affects its rotational inertia. The bent rod exhibits dynamics different from a linear rod because of its shifted symmetry and two perpendicular sections. To predict how various forces influence its rotation, we calculate moments of inertia, considering the shape's complexity. Knowledge of this field helps solve problems related to machinery parts, leverage tools, and even celestial motion. For this rod, when evaluating rotational dynamics, dividing the rod into two parts provides a systematic approach. Each segment is assessed for how it contributes to the overall inertia. Such assessments lead to better understanding of stability and balance in rotating systems.

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Most popular questions from this chapter

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius 25.0 cm. Starting from rest at \(t =\) 0, the flywheel rotates with constant angular acceleration 3.00 rad/s\(^2\) about an axis perpendicular to the flywheel at its center. If the flywheel has a density (mass per unit volume) of 8600 kg/m\(^3\), what thickness must it have to store 800 J of kinetic energy at \(t =\) 8.00 s?

As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be 0.640 m. Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an 8.20-kg block of wood from the free end of the rope, as in Fig. E9.45. You release the system from rest and find that the block descends 12.0 m in 4.00 s. What is the moment of inertia of the wheel for this axis?

You need to design an industrial turntable that is 60.0 cm in diameter and has a kinetic energy of 0.250 J when turning at 45.0 rpm 1rev/min2. (a) What must be the moment of inertia of the turntable about the rotation axis? (b) If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?

A thin, uniform rod is bent into a square of side length \(a\). If the total mass is \(M\), find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (\(Hint\): Use the parallel-axis theorem.)

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant \(linear\) speed of \(v =\) 1.25m/s. Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the CD is played. (See Exercise 9.20.) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta) = r_0 + \beta\theta\), where \(r_0\) is the radius of the spiral at \(\theta =\) 0 and \(\beta\) is a constant. On a CD, \(r_0\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d\theta\), the distance scanned along the track is \(ds = rd\theta\). Using the above expression for \(r(\theta)\), integrate \(ds\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) Since the track is scanned at a constant linear speed \(v\), the distance s found in part (a) is equal to \(vt\). Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta\) ; choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_z\) and the angular acceleration \(\alpha_z\) as functions of time. Is \(\alpha_z\) constant? (d) On a CD, the inner radius of the track is 25.0 mm, the track radius increases by 1.55 mm per revolution, and the playing time is 74.0 min. Find \(r_0, \beta,\) and the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_z\) (in rad/s) versus \(t\) and \(\alpha_z\) (in rad/s\(^2\)) versus \(t\) between \(t =\) 0 and \(t =\) 74.0 min. \(\textbf{The Spinning eel.}\) American eels (\(Anguilla\) \(rostrata\)) are freshwater fish with long, slender bodies that we can treat as uniform cylinders 1.0 m long and 10 cm in diameter. An eel compensates for its small jaw and teeth by holding onto prey with its mouth and then rapidly spinning its body around its long axis to tear off a piece of flesh. Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way. Although this feeding method is costly in terms of energy, it allows the eel to feed on larger prey than it otherwise could.

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