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In redesigning a piece of equipment, you need to replace a solid spherical part of mass \(M\) with a hollow spherical shell of the same size. If both parts must spin at the same rate about an axis through their center, and the new part must have the same kinetic energy as the old one, what must be the mass of the new part in terms of \(M ?\)

Short Answer

Expert verified
The mass of the hollow sphere must be \( \frac{3}{5}M \).

Step by step solution

01

Understanding Kinetic Energy in Rotation

The kinetic energy due to rotation for any object is given by \[ KE = \frac{1}{2}I\omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For two objects to have the same kinetic energy with the same angular velocity, their moments of inertia about the axis must be equal.
02

Moment of Inertia of a Solid Sphere

The moment of inertia for a solid sphere rotating about its diameter is \[ I_{solid} = \frac{2}{5}MR^2 \] where \( M \) is the mass and \( R \) is the radius.
03

Moment of Inertia of a Hollow Sphere

The moment of inertia for a hollow sphere rotating about its diameter is \[ I_{hollow} = \frac{2}{3}mR^2 \] where \( m \) is the mass of the hollow sphere and \( R \) is its radius. Since the hollow shell and the solid sphere have the same size, \( R \) is the same for both.
04

Equating the Moments of Inertia

Since both spheres must have the same kinetic energy with identical angular velocity, their moments of inertia must be equal: \[ I_{solid} = I_{hollow} \]Thus, \[ \frac{2}{5}MR^2 = \frac{2}{3}mR^2 \]
05

Solving for Mass of the Hollow Sphere

Cancel \( R^2 \) on both sides of the equation and solve for \( m \):\[ \frac{2}{5}M = \frac{2}{3}m \]Multiply through by 15 to clear fractions:\[ 6M = 10m \]Divide by 10:\[ m = \frac{3}{5}M \]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In the context of rotational motion, this energy is captured by the formula:
  • \[ KE = \frac{1}{2}I\omega^2 \] where \( KE \) is the kinetic energy, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
For rotating objects like spheres, it is essential to understand that kinetic energy ties closely with how mass is distributed. A solid sphere with mass \(M\) will have a different kinetic energy compared to a hollow sphere with mass \(m\) if their shapes and sizes differ. However, if their angular velocities and energy levels must match, we need to adjust their masses accordingly. This is pivotal in engineering, where balancing energy outputs without changing speeds can prevent machinery problems.
In our example, to maintain the same kinetic energy with a switch from a solid to a hollow sphere, we adjusted the mass of the hollow sphere to keep their energy outputs equivalent.
Moment of Inertia
The moment of inertia, denoted as \( I \), is a property of an object that quantifies its resistance to rotational motion. It varies with the mass distribution relative to the axis of rotation. For a solid sphere, it is calculated as:
  • \[ I_{solid} = \frac{2}{5}MR^2 \]
  • For a hollow sphere, the formula is slightly different: \[ I_{hollow} = \frac{2}{3}mR^2 \]
Here, \( M \) and \( m \) represent the masses of the solid and hollow spheres respectively, while \( R \) is the radius, which remains consistent in our redesign scenario. The moment of inertia is crucial in physics because it affects how much torque is needed to achieve a desired angular acceleration. In this problem, matching the kinetic energy required us to equate these two moments of inertia, which led to solving for the mass of the hollow sphere. This balance allows systems in rotational dynamics to function smoothly without unsettling the overall mechanical motion.
Rotational Dynamics
Rotational dynamics deals with the motion of objects that are rotating or spinning and involves complex interactions between forces and moments. Key elements such as angular velocity, torque, and moment of inertia all interact within this field. When addressing a redesign of machinery with rotating parts like spheres, understanding these interactions is crucial.
  • The angular velocity \( \omega \) is vital because it retains similarity in both spheres as required in our scenario.
  • Equally important is achieving equal kinetic energy offering a seamless transition in machinery operations despite the change in component type.
In engineering terms, designing equipment that switches from solid to hollow spheres while keeping rotational dynamics consistent could enhance performance, providing lighter machines with the same efficiency. This process involves balancing the angular properties and mass distribution to prevent any negative impacts on machinery operation. By considering all these aspects, it embodies the practical application of physics in real-world problem-solving, ensuring that the equipment remains functional and efficient post-redesign.

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Most popular questions from this chapter

\(\bullet\) (a) A cylinder 0.150 \(\mathrm{m}\) in diameter rotates in a lathe at 620 \(\mathrm{rpm} .\) What is the tangential speed of the surface of the cylinder? (b) The proper tangential speed for machining cast iron is about 0.600 \(\mathrm{m} / \mathrm{s} .\) At how many revolutions per minute should a piece of stock 0.0800 \(\mathrm{m}\) in diameter be rotated in a lathe to produce this tangential speed?

\(\bullet\) A flywheel with a radius of 0.300 \(\mathrm{m}\) starts from rest and accelerates with a constant angular acceleration of 0.600 \(\mathrm{rad} / \mathrm{s}^{2} .\) Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim (a) at the start, (b) after it has turned through \(60.0^{\circ},\) and \((\mathrm{c})\) after it has turned through \(120.0^{\circ} .\)

\(\bullet\) An airplane propeller is rotating at 1900 \(\mathrm{rpm.}\) (a) Compute the propeller's angular velocity in rad/s. (b) How many seconds does it take for the propeller to turn through \(35^{\circ} ?\) (c) If the propeller were turning at \(18 \mathrm{rad} / \mathrm{s},\) at how many rpm would it be turning? (d) What is the period (in seconds) of this propeller?

A compound disk of outside diameter 140.0 \(\mathrm{cm}\) is made up of a uniform solid disk of radius 50.0 \(\mathrm{cm}\) and area density 3.00 \(\mathrm{g} / \mathrm{cm}^{2}\) surrounded by a concentric ring of inner radius \(50.0 \mathrm{cm},\) outer radius \(70.0 \mathrm{cm},\) and area density 2.00 \(\mathrm{g} / \mathrm{cm}^{2} .\) Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center.

\(\bullet\) When a toy car is rapidly scooted across the floor, it stores energy in a flywheel. The car has mass \(0.180 \mathrm{kg},\) and its fly- wheel has moment of inertia \(4.00 \times 10^{-5} \mathrm{kg} \cdot \mathrm{m}^{2} .\) The car is 15.0 \(\mathrm{cm}\) long. An advertisement claims that the car can travel at a scale speed of up to 700 \(\mathrm{km} / \mathrm{h}(440 \mathrm{mi} / \mathrm{h}) .\) The scale speed is the speed of the toy car multiplied by the ratio of the length of an actual car to the length of the toy. Assume a length of 3.0 \(\mathrm{m}\) for a real car. (a) For a scale speed of \(700 \mathrm{km} / \mathrm{h},\) what is the actual translational speed of the car? (b) If all the kinetic energy that is initially in the flywheel is converted to the translational kinetic energy of the toy, how much energy is originally stored in the flywheel? (c) What ini- tial angular velocity of the flywheel was needed to store the amount of energy calculated in part (b)?

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