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Chapter 8: Problem 31

(I) Estimate the moment of inertia of a bicycle wheel 67 cm in diameter. The rim and tire have a combined mass of 1.1 kg. The mass of the hub (at the center) can be ignored (why?).

Short Answer

Expert verified
The moment of inertia of the bicycle wheel is approximately 0.123 kg m^2.

Step by step solution

01

Understanding the Problem

We need to calculate the moment of inertia (I) for a bicycle wheel with its mass concentrated at the rim. The bicycle wheel's diameter is 67 cm, and the mass of the rim and tire is 1.1 kg. The hub's mass is ignored because its contribution to the moment of inertia is negligible compared to the rim's mass, as the hub is at the center.
02

Convert Diameter to Radius

The diameter of the wheel is given as 67 cm. To find the radius (r), divide the diameter by 2: \( r = \frac{67}{2} = 33.5 \text{ cm} \). Convert this to meters as well: \( r = 0.335 \text{ m} \).
03

Use the Moment of Inertia Formula

For a thin hoop or ring, which the wheel can be modeled as, the moment of inertia is given by \( I = m \, r^2 \), where \( m \) is the mass and \( r \) is the radius.
04

Substitute Known Values

We have \( m = 1.1 \text{ kg} \) and \( r = 0.335 \text{ m} \). Substitute these values into the formula: \( I = 1.1 \, \times \, (0.335)^2 \).
05

Calculate the Moment of Inertia

Calculate \( I = 1.1 \, \times \, (0.335)^2 \approx 1.1 \, \times \, 0.112225 = 0.1234475 \text{ kg} \, \text{m}^2 \).

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Key Concepts

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

Bicycle Wheel
When dealing with a bicycle wheel, it's important to understand its structure and how it affects its moment of inertia. A bicycle wheel generally consists of a rim and a hub. The rim holds the tire and is located at the wheel's edge, while the hub is at the center, connecting to the bicycle frame. For exercises involving physics calculations, like calculating the moment of inertia, we often focus on the rim. This is because the rim holds most of the wheel's mass, notably affecting the wheel鈥檚 rotational inertia. Since the hub is at the center and carries minimal mass, it doesn't significantly contribute to the moment of inertia, allowing us to ignore it in such calculations.
Thin Hoop Approximation
Modeling a bicycle wheel as a thin hoop simplifies the calculation of its moment of inertia. The 'thin hoop approximation' assumes that all the mass of the wheel is concentrated at the rim, creating a hoop or ring shape.
This makes it easier to apply the formula for the moment of inertia:
  • The moment of inertia for a thin hoop is expressed as \( I = m \, r^2 \).
  • Here, \( m \) is the mass of the hoop (rim), and \( r \) is the radius of the wheel.
This formula is derived from the idea that the mass is evenly distributed around a circular path, with each point contributing to rotation equally. The thin hoop approximation is particularly helpful for objects like bicycle wheels since most of the mass resides at the rim, much like a hoop.
Mass and Radius Calculation
To calculate the moment of inertia, it's crucial to determine both the mass and radius of the bicycle wheel correctly. Here鈥檚 how we do it:
  • **Mass Calculation:** In our exercise, the combined mass of the rim and tire is provided as 1.1 kg. There鈥檚 no need to include the hub since its mass impact is negligible.
  • **Radius Calculation:** The diameter of the wheel is given as 67 cm. To find the radius, simply divide the diameter by 2. Convert this number into meters for scientific calculations: \( r = \frac{67}{2} = 33.5 \text{ cm} = 0.335 \text{ m} \).
With these values, you can substitute them into the moment of inertia formula \( I = m \, r^2 \). This gives you a clear path to find the inertia considering the simplified thin hoop model.

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Most popular questions from this chapter

(I) What is the angular momentum of a 0.270-kg ball revolving on the end of a thin string in a circle of radius 1.35 m at an angular speed of 10.4 rad/s?

(II) A potter's wheel is rotating around a vertical axis through its center at a frequency of 1.5 rev/s. The wheel can be considered a uniform disk of mass 5.0 kg and diameter 0.40 m. The potter then throws a 2.6-kg chunk of clay, approximately shaped as a flat disk of radius 7.0 cm, onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it? Ignore friction.

(II) A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.0 m and moment of inertia \(820 kg\cdot m^2\). The platform rotates without friction with angular velocity 0.95 rad/s. The person walks radially to the edge of the platform. \((a)\) Calculate the angular velocity when the person reaches the edge. \((b)\) Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.

A merry-go-round with a moment of inertia equal to \(1260 kg\cdot m^2\) and a radius of 2.5 m rotates with negligible friction at 1.70 rad/s. A child initially standing still next to the merry-go-round jumps onto the edge of the platform straight toward the axis of rotation, causing the platform to slow to 1.35 rad/s. What is her mass?

(II) Let us treat a helicopter rotor blade as a long thin rod, as shown in Fig. 8-49. \((a)\) If each of the three rotor helicopter blades is 3.75 m long and has a mass of 135 kg, calculate the moment of inertia of the three rotor blades about the axis of rotation. \((b)\) How much torque must the motor apply to bring the blades from rest up to a speed of 6.0 rev/s in 8.0 s?
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