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A metal sign for a car dealership is a thin, uniform right triangle with base length \(b\) and height \(h\) . The sign has mass \(M\) . (a) What is the moment of inertia of the sign for rotation about the side of length \(h ?\) If \(M=5.40 \mathrm{kg}, b=1.60 \mathrm{m},\) and \(h=1.20 \mathrm{m},\) what is the kinetic energy of the sign when it is rotating about an axis along the \(1.20-\mathrm{m}\) side at 2.00 \(\mathrm{rev} / \mathrm{s} ?\)

Short Answer

Expert verified
The moment of inertia is 4.608 kg⋅m², and the kinetic energy is approximately 364.78 J.

Step by step solution

01

Calculating the Moment of Inertia

The moment of inertia for a uniform right triangle rotating about one of its sides can be calculated using the formula for a right triangle, which is: \[ I = \frac{1}{3} M b^2 \] where \( M \) is the mass of the triangle, and \( b \) is the base of the triangle. Here, \( M = 5.40\, \text{kg} \) and \( b = 1.60\, \text{m} \). Substitute the given values: \[ I = \frac{1}{3} \times 5.40\, \text{kg} \times (1.60\, \text{m})^2 = \frac{1}{3} \times 5.40 \times 2.56 = 4.608\, \text{kg} \cdot \text{m}^2 \].
02

Convert Rotational Speed to Radians per Second

Since the rotational speed is given in revolutions per second, it can be converted to units of radians per second by using the conversion factor \(1\, \text{rev} = 2\pi\, \text{rad}\). So, if the rotational speed is \(2.00\, \text{rev/s}\), the corresponding rotational speed in radians per second is \[ \omega = 2.00 \times 2\pi = 4\pi \approx 12.57\, \text{rad/s} \].
03

Calculating the Kinetic Energy

The kinetic energy of a rotating object can be calculated using the formula: \[ KE = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. Substitute \( I = 4.608\, \text{kg}\cdot\text{m}^2 \) and \( \omega = 12.57\, \text{rad/s} \): \[ KE = \frac{1}{2} \times 4.608 \times (12.57)^2 \approx 364.78\, \text{J} \].

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Key Concepts

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

Kinetic Energy
Kinetic energy represents the energy that an object possesses due to its motion. For rotational motion, the formula to calculate kinetic energy is slightly different from that of linear motion. In rotational systems, we use the formula: \[ KE = \frac{1}{2} I \omega^2 \]This equation depends on two main factors: the moment of inertia \( I \) and the angular velocity \( \omega \). The moment of inertia measures how the mass is distributed relative to the axis of rotation. The angular velocity is the rate at which the object spins, measured in radians per second.
- Moment of inertia is comparable to mass in linear motion and affects how easily an object can start or stop rotating. - Angular velocity indicates the speed of rotation.
In our exercise, we calculated a kinetic energy of approximately 364.78 Joules for a rotating right triangle sign, meaning it uses that amount of energy to maintain its rotational speed.
Right Triangle
A right triangle is a specific type of triangle with one angle measuring exactly 90 degrees. This triangle has several unique properties that make it crucial in calculations related to geometry and physics. - The two sides forming the right angle are called the 'legs', and the side opposite the right angle is the 'hypotenuse'.
In our scenario, the metal sign is a right triangle with a base \( b \) and a height \( h \). For calculating the moment of inertia, only the base and the mass of the triangle are needed using the formula \[ I = \frac{1}{3} M b^2 \]This indicates that the sign rotates about one of its legs, treating the base as the axis. When considering rotational motion, the structure and proportions of a right triangle critically influence the computation of rotational attributes.
Rotational Motion
Rotational motion occurs when an object spins around an axis. It is a fundamental concept in physics involving parameters like moment of inertia, angular velocity, and angular acceleration.
The motion of a body rotating about an axis involves different calculations compared to objects moving linearly:- **Angular Velocity (\( \omega \))**: It measures how fast an object rotates or revolves relative to another point, expressed in radians per second.- **Moment of Inertia (\( I \))**: This gives an idea of how difficult it is to change the rotation of an object. It is dependent on the mass distribution of the object.
In our scenario, the right triangle is rotating about its side, and its angular velocity was given in revolutions per second. We converted this to radians per second using \( 1 \text{ rev} = 2\pi \text{ rad} \).The connection between these elements affects both the angular momentum and the kinetic energy of the rotating object. More mass farther from the center of rotation increases the moment of inertia, making the object harder to spin or stop.

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Most popular questions from this chapter

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

A bicycle wheel has an initial angular velocity of 1.50 \(\mathrm{rad} / \mathrm{s}\) . (a) If its angular acceleration is constant and equal to 0.300 \(\mathrm{rad} / \mathrm{s}^{2}\) , what is its angular velocity at \(t=2.50 \mathrm{s} ?\) (b) Through what angle has the wheel turned between \(t=0\) and \(t=2.50 \mathrm{s} ?\)

A thin uniform rod 50.0 \(\mathrm{cm}\) long with mass 0.320 \(\mathrm{kg}\) is bent at its center into a \(\mathrm{V}\) shape, with a \(70.0^{\circ}\) angle at its vertex. Find the moment of inertia of this \(\mathrm{V}\) -shaped object about an axis perpendicular to the plane of the \(\mathrm{V}\) at its vertex.

Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits \(10^{-7} \mathrm{m}\) deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 \(\mathrm{mm}\) and 58.0 \(\mathrm{mm}\) , respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 \(\mathrm{m} / \mathrm{s}\) . (a) What is the angular speed of the CD when the innermost part of the track is scanned? The outermost part of the track? (b) The maximum playing time of a \(\mathrm{CD}\) is 74.0 min. What would be the length of the track on such a maximum-duration \(\mathrm{CD}\) if it were stretched out in a straight line? (c) What is the average angular acceleration of a maximum-duration CD during its 74.0 -min playing time? Take the direction of rotation of the dise to be positive.

About what axis will a uniform, balsa-wood sphere have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius, with the axis along a diameter?

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