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How I Scales. If we multiply all the design dimensions of an object by a scaling factor \(f\), its volume and mass will be multiplied by \(f ^3\). (a) By what factor will its moment of inertia be multiplied? (b) If a \(\frac{1}{48}\)-scale model has a rotational kinetic energy of 2.5 J, what will be the kinetic energy for the full-scale object of the same material rotating at the same angular velocity?

Short Answer

Expert verified
(a) The moment of inertia is multiplied by \( f^5 \). (b) The full-scale kinetic energy is 637,009,920 J.

Step by step solution

01

Understanding Scaling of Moment of Inertia

The moment of inertia depends on the distribution of mass, thus when dimensions scale by factor \( f \), all linear dimensions scale linearly by \( f \). The volume scales by \( f^3 \), and hence mass also scales by \( f^3 \), given a constant density.The moment of inertia for a point mass is \( I = mr^2 \), which means for scaled objects, the radius scales by \( f \). Therefore, considering both mass and radius influence, the moment of inertia scales by \( f^3 \times f^2 = f^5 \).
02

Applying the Scaling Factor for Moment of Inertia

Given that dimensions scale by a factor of \( f = 48 \) for a full-scale object, the moment of inertia of the full-scale object will be the model's moment of inertia multiplied by \( 48^5 \).
03

Finding Kinetic Energy Scaling Factor

Rotational kinetic energy is given by \( K = \frac{1}{2} I \omega^2 \). If \( \omega \) is kept constant, only the scaling of moment of inertia affects the kinetic energy. Thus, if the moment of inertia is scaled by \( 48^5 \), the kinetic energy will also be scaled by \( 48^5 \).
04

Calculating Full-Scale Kinetic Energy

The kinetic energy of the model is 2.5 J. To find the full-scale kinetic energy, multiply this by the scaling factor: \( 2.5 \times 48^5 \). First, calculate \( 48^5 = 254803968 \). Then, the full-scale kinetic energy is \( 2.5 \times 254803968 = 637009920 \) J.

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Key Concepts

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

Scaling Factor
When we talk about scaling factors, we're referring to how much larger or smaller an object becomes when we change its size. Imagine taking a small model and making it into a full-sized object. If all dimensions of the model are increased by a factor of "f," the mass and volume of the object will be scaled by a factor of \( f^3 \). This is because volume and mass are three-dimensional properties.The moment of inertia, which depends on how mass is distributed within the object, also scales based on both the mass and the radius. For a point mass, the moment of inertia is \( I = mr^2 \). So, if the object is scaled by "f," mass scales by \( f^3 \) and radius by "f." Therefore, the moment of inertia scales by \( f^5 \), multiplying both mass and radius changes.Understanding these scaling concepts is crucial when moving from models to real-life objects, helping in calculating different physical properties accurately.
Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its motion around an axis. If you imagine a spinning top or a rotating wheel, that's rotational kinetic energy in action. The formula for calculating this energy is:\[ K = \frac{1}{2} I \omega^2 \]where:- \( K \) is the rotational kinetic energy.- \( I \) is the moment of inertia.- \( \omega \) is the angular velocity.In the context of scaling, if you scale up an object while keeping the angular velocity constant, the only thing that changes the kinetic energy is the moment of inertia. Hence, the change in kinetic energy is directly tied to how the moment of inertia scales. For example, in our case where a small model has a kinetic energy of 2.5 J, the full-scale object has its moment of inertia and, therefore, its kinetic energy increased by the factor \( 48^5 \). This shows how scaling can drastically change the energy required or produced by an object when it rotates.
Angular Velocity
Angular velocity represents how quickly an object rotates or spins. It tells you how fast the angle is changing as the object moves around a point or axis.Just like how speed measures distance over time, angular velocity measures the angle over time. It usually has the symbol \( \omega \) and is measured in radians per second.In scaling scenarios, if you maintain the same angular velocity for both small models and full-sized objects, the speed at which they rotate remains the same. Therefore, the angular velocity does not change as we scale the size of the object.Consequently, in calculations involving rotational kinetic energy, as long as the angular velocity is kept constant (\( \omega \) does not change), the change in kinetic energy when scaling is only dependent on how the moment of inertia changes. This is key when trying to understand the physical behaviors of objects at different sizes while maintaining similar rotational dynamics.

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Most popular questions from this chapter

A circular saw blade 0.200 m in diameter starts from rest. In 6.00 s it accelerates with constant angular acceleration to an angular velocity of 140 rad/s. Find the angular acceleration and the angle through which the blade has turned.

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s\(^2\). Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) \(a_{rad} = \omega^2r\) and (b) \(a_{rad} = v^2/r\)

The angular velocity of a flywheel obeys the equation \(\omega_z\)(\(t\)) \(= A + Bt^2\), where \(t\) is in seconds and \(A\) and \(B\) are constants having numerical values 2.75 (for \(A\)) and 1.50 (for \(B\)). (a) What are the units of \(A\) and \(B\) if \(\omega_z\) is in rad/s? (b) What is the angular acceleration of the wheel at (i) \(t = 0\) and (ii) \(t =\) 5.00 s? (c) Through what angle does the flywheel turn during the first 2.00 s? (\(Hint\): See Section 2.6.)

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.)

A sphere with radius \(R =\) 0.200 m has density \(\rho\) that decreases with distance \(r\) from the center of the sphere according to \(r =\) 3.00 \(\times\) 103 kg/m\(^3 -\) (9.00 \(\times\) 103 kg/m\(^4\))\(r\). (a) Calculate the total mass of the sphere. (b) Calculate the moment of inertia of the sphere for an axis along a diameter.

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