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(II) A child rolls a ball on a level floor 3.5 m to another child. If the ball makes 12.0 revolutions, what is its diameter?

Short Answer

Expert verified
The ball's diameter is approximately 0.0927 meters.

Step by step solution

01

Understand the Problem

We need to find the diameter of the ball. We know the ball rolls 3.5 meters and makes 12.0 revolutions.
02

Formula for Distance Covered

The distance covered by the ball when it rolls is related to the number of revolutions it makes and its circumference. This relationship can be expressed as: \[ \text{Distance} = \text{Revolutions} \times \text{Circumference} \] where \( \text{Circumference} = \pi \times \text{Diameter} \).
03

Plug Values into the Formula

Given that the distance is 3.5 meters and the ball makes 12 revolutions:\[ 3.5 = 12 \times (\pi \times \text{Diameter}) \].
04

Solve for Diameter

Rearrange the equation to solve for the diameter:\[ \text{Diameter} = \frac{3.5}{12 \times \pi} \].
05

Calculate the Diameter

Perform the calculation:\[ \text{Diameter} = \frac{3.5}{12 \times 3.1416} \approx 0.0927 \text{ meters} \].

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Key Concepts

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

Distance Calculation
Understanding how to calculate distance is crucial in many physics problems, especially when dealing with motion. In this exercise, we are given that a ball travels a certain distance on the floor. The key idea is that distance is not only about straight lines but involves understanding the object's path. Here, the ball's path is circular since it's making revolutions.
The distance travelled by a rolling object can be calculated using the formula:
  • \( \text{Distance} = \text{Revolutions} \times \text{Circumference} \)
Think of it like unrolling the ball's path in straight lines. Each revolution signifies one complete trip around the ball's circumference. So multiplying the number of revolutions by the circumference gives us the total distance the ball has traveled linearly. Keep in mind that seeing the movement this way helps in solving many such physics problems.
Circumference Formula
The circumference formula is foundational for figuring out the distance covered in scenarios involving circular motion. When objects like wheels or balls roll, knowing the circumference is key to understanding how far they travel in one full revolution.
The formula to calculate the circumference of a circle is:
  • \( \text{Circumference} = \pi \times \text{Diameter} \)
Here, \( \pi \) (Pi) is approximately 3.1416, a mathematical constant representing the ratio of a circle’s circumference to its diameter. Using this formula, you can relate the diameter of a spherical object to its circumference. This relationship is a fundamental concept when determining how the size of an object affects its rolling distance. By rearranging this formula, you can solve for the diameter if you know the circumference, adding a powerful tool to your problem-solving toolbox.
Revolutions and Motion
Revolutions are a way to describe circular motion quantitatively. When dealing with objects that spin or roll, revolutions tell us how many complete turns the object makes. Each revolution is equivalent to the object traveling its entire circumference once.
Using revolutions to solve problems:
  • When given the number of revolutions, you can multiply by the circumference to find the total distance traveled.
  • It bridges the gap between rotational and linear movement by linking them through circumference.
In our exercise, 12 revolutions indicate that the ball has covered the distance of 12 circumferences. By using this approach, you're able to calculate not only how far it rolled but can also deduce information about the object’s size. This concept is incredibly useful in various applications, from engineering to everyday problem-solving. Understanding this can help you tackle more complex questions that involve rotational dynamics or mechanics.

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Most popular questions from this chapter

(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?

(II) A potter is shaping a bowl on a potter's wheel rotating at constant angular velocity of 1.6 rev/s (Fig. 8-48). The friction force between her hands and the clay is 1.5 N total. \((a)\) How large is her torque on the wheel, if the diameter of the bowl is 9.0 cm? \((b)\) How long would it take for the potter's wheel to stop if the only torque acting on it is due to the potter's hands? The moment of inertia of the wheel and the bowl is \(0.11 kg\cdot m^2\).

(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?

(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?).

(II) A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The small wheel has a radius of 2.0 cm and accelerates at the rate of \(7.2 rad/s^2\), and it is in contact with the pottery wheel (radius 27.0 cm) without slipping. Calculate \((a)\) the angular acceleration of the pottery wheel, and \((b)\) the time it takes the pottery wheel to reach its required speed of 65 rpm.

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