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A semicircular panel with a radius r is attached with hinges to a circular plate with a radius r and initially is held in the vertical position as shown. The plate and the panel are made of the same material and have the same thickness. Knowing that the entire assembly is rotating freely with an initial angular velocity of \(\omega_{0}\), determine the angular velocity of the assembly after the panel has been released and comes to rest against the plate.

Short Answer

Expert verified
The angular velocity after the panel falls is \(\frac{2}{7}\omega_0\).

Step by step solution

01

Identify the System

We have a semicircular panel attached to a circular plate. Both components have the same radius, \(r\), are made from the same material, and have the same thickness. This means they also have the same density.
02

Define the Moment of Inertia

The moment of inertia for the circular plate (disk) about its center is \(I_{disk} = \frac{1}{2} mr^2\), where \(m\) is the mass of the disk. For the semicircular panel, which is like a half-disk, the moment of inertia about the center of the full circle is \(I_{panel} = \frac{1}{2} \times \frac{1}{2} mr^2 = \frac{1}{4}mr^2\). Since the panel rotates about the hinges at its edge, we use the parallel axis theorem to find the panel's moment of inertia about this axis as \(I_{panel} = \frac{1}{4}mr^2 + mr^2 = \frac{5}{4}mr^2\).
03

Total Initial and Final Moments of Inertia

Initially, the assembly's moment of inertia is just that of the disk, \(I_{initial} = \frac{1}{2}mr^2\), since the panel starts in the vertical position and contributes negligibly to the initial rotational inertia. After the panel has been released and lies flat, the total moment of inertia is \(I_{final} = I_{disk} + I_{panel} = \frac{1}{2}mr^2 + \frac{5}{4}mr^2 = \frac{7}{4}mr^2\).
04

Apply Conservation of Angular Momentum

Since there are no external torques acting on the system, we apply the conservation of angular momentum. Initially, the assembly is rotating with angular velocity \(\omega_0\). Therefore, \(I_{initial} \times \omega_0 = I_{final} \times \omega_f\). Substitute the moments of inertia: \(\frac{1}{2}mr^2 \cdot \omega_0 = \frac{7}{4}mr^2 \cdot \omega_f\). Simplifying gives the angular velocity \(\omega_f = \frac{2}{7}\omega_0\).
05

Calculate Final Angular Velocity

Solve the equation from Step 4 for \(\omega_f\):\[ \omega_f = \frac{2}{7} \omega_0 \]This is the angular velocity of the entire assembly after the panel lies flat against the plate.

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Key Concepts

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

moment of inertia
Imagine trying to make a heavy merry-go-round spin. It's tougher than spinning a light bicycle wheel, isn't it? That difficulty is due to a concept called moment of inertia. In simple terms, the moment of inertia measures how much an object resists changes in its rotational motion. The heavier or more widely distributed an object's mass is, the greater its moment of inertia.
For a circular disk, like the one in the exercise, the moment of inertia around its center is given by the formula: \[ I_{disk} = \frac{1}{2}mr^2 \] where \( m \) is the mass of the disk and \( r \) is the radius. The semicircular panel’s moment of inertia is different because of its shape and position of rotation. Initially, it has a smaller contribution, but eventually, we determine its full contribution by accounting for its pivot point using the parallel axis theorem. This advanced calculation helps describe how the entire assembly balances its rotational resistance when the shape changes as the panel falls away. Understanding these principles is crucial in the physics of rotational systems.
conservation of angular momentum
Ever spin around holding weights and feel the pull as soon as you open your arms? That's an example of the conservation of angular momentum. In the simplified world of physics, if nothing external like friction stops you, your rate of spinning times your moment of inertia stays constant.
In the problem, the rotating plate and panel experience this conservation principle. Initially, the system, just the disk, spins with an angular velocity of \( \omega_0 \). As the panel descends, altering the moment of inertia, the total spin must adjust to keep momentum constant. In equation form:\[ I_{initial} \times \omega_0 = I_{final} \times \omega_f \]The absence of external torques means every change in shape (like the panel dropping) adjusts the velocity (how fast it spins) inversely. After substituting with precise inertia calculations, the final velocity can be determined, demonstrating the elegant balance nature holds in rotational motion.
rotational dynamics
To see rotational dynamics in action, imagine different ways objects spin. It's not just about how fast, like a CD that whirls at speed, but involves how forces and motions distribute across the spinning object. Rotational dynamics describes how torques (twisting forces) and angular motions form and balance, much like linear dynamics in straight-line motions.
In the problem, when the semicircular panel falls, it impacts the whole system's rotational attributes. Initially, the rotational dynamics of only the disk matter, but as the panel finds its new position, physics rules of torque and inertia dictate how both parts transmit and conserve motion. These principles help explain engineering challenges, from balancing tires to spacecraft navigation. Every rotational system obeys these fundamental principles, connecting local moments to the broader spin and stability of the object.
Learning these concepts equips students with an intuitive understanding of how real-world spinning systems function.

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Most popular questions from this chapter

The 40 -kg gymnast drops from her maximum height of \(h=0.5 \mathrm{m}\) straight down to the bar as shown. Her hands hit the bar and clasp onto it, and her body remains straight in the position shown. Her center of mass is 0.75 meters away from her hands, and her mass moment of inertia about her center of mass is \(7.5 \mathrm{kg} \cdot \mathrm{m}^{2}\). Assuming that friction between the bar and her hands is negligible and that she remains in the same position throughout the swing, determine her angular velocity when she swings around to \(\theta=135^{\circ} .\)

Sphere \(A\) of mass \(m\) and radius \(r\) rolls without slipping with a velocity \(\bar{v}_{1}\) on a horizontal surface when it hits squarely an identical sphere \(B\) that is at rest. Denoting by \(\mu_{k}\) the coefficient of kinetic friction between the spheres and the surface, neglecting friction between the spheres, and assuming perfectly elastic impact, determine \((a)\) the linear and angular velocities of each sphere immediately after the impact, (b) the velocity of each sphere after it has started rolling uniformly.

The flywheel of a small punch rotates at 300 rpm. It is known that 1800 ft?lb of work must be done each time a hole is punched. It is desired that the speed of the flywheel after one punching be not less than 90 percent of the original speed of 300 rpm. (a) Determine the required moment of inertia of the flywheel. (b) If a constant 25-lb?ft couple is applied to the shaft of the flywheel, determine the number of revolutions that must occur between each punching, knowing that the initial velocity is to be 300 rpm at the start of each punching.

Sphere \(A\) of mass \(m_{A}=2 \mathrm{kg}\) and radius \(r=40 \mathrm{mm}\) rolls without slipping with a velocity \(\bar{v}_{1}=2 \mathrm{m} / \mathrm{s}\) on a horizontal surface when it hits squarely a uniform slender bar \(B\) of mass \(m_{B}=0.5 \mathrm{kg}\) and length \(L=100 \mathrm{mm}\) that is standing on end and is at rest. Denoting by \(\mu_{k}\) the coefficient of kinetic friction between the sphere and the horizontal surface, neglecting friction between the sphere and the bar, and knowing the coefficient of restitution between A and B is 0.1, determine the angular velocities of the sphere and the bar immediately after the impact.

Greek engineers had the unenviable task of moving large columns from the quarries to the city. One engineer, Chersiphron, tried several different techniques to do this. One method was to cut pivot holes into the ends of the stone and then use oxen to pull the column. The 4-ft diameter column weighs 12,000 lbs, and the team of oxen generates a constant pull force of 1500 lbs on the center of the cylinder G. Knowing that the column starts from rest and rolls without slipping, determine (a) the velocity of its center G after it has moved 5 ft, (b) the minimum static coefficient of friction that will keep it from slipping.

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