/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 (II) A softball player swings a ... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) A softball player swings a bat, accelerating it from rest to 2.7 \(\mathrm{rev} / \mathrm{s}\) in a time of 0.20 \(\mathrm{s}\) . Approximate the bat as a 2.2 -kg uniform rod of length \(0.95 \mathrm{m},\) and compute the torque the player applies to one end of it.

Short Answer

Expert verified
The torque is approximately 56.29 N·m.

Step by step solution

01

Determining the Moment of Inertia

To calculate the torque, we first need to determine the moment of inertia (I) of the bat. Approximating the bat as a uniform rod rotating about one end, the moment of inertia is given by the formula: \[ I = \frac{1}{3} m L^2 \]where \(m\) is the mass of the rod (2.2 kg) and \(L\) is the length of the rod (0.95 m). Substituting these values, we get:\[ I = \frac{1}{3} \times 2.2 \times (0.95)^2 = 0.664 \: \text{kg} \cdot \text{m}^2 \]
02

Calculating Angular Acceleration

Next, we need to find the angular acceleration (\(\alpha\)). The angular acceleration is defined as the change in angular velocity over time. The initial angular velocity (\(\omega_i\)) is 0, and the final angular velocity (\(\omega_f\)) is 2.7 rev/s, which we convert to radians per second:\[ \omega_f = 2.7 \cdot 2\pi \approx 16.96 \: \text{rad/s} \] The time given is \(t = 0.20\) s. Thus, \(\alpha\) is:\[ \alpha = \frac{\omega_f - \omega_i}{t} = \frac{16.96 - 0}{0.20} = 84.8 \: \text{rad/s}^2 \]
03

Calculating the Torque

Finally, we can calculate the torque (\(\tau\)) using the equation:\[ \tau = I \cdot \alpha \]Substituting the values of \(I = 0.664\) kg·m² and \(\alpha = 84.8\) rad/s², we get:\[ \tau = 0.664 \cdot 84.8 = 56.29 \: \text{N} \cdot \text{m} \]
04

Conclusion

The torque applied by the softball player to the bat is approximately \(56.29 \: \text{N} \cdot \text{m}\).

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Key Concepts

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

Moment of Inertia
When dealing with rotational motion, the moment of inertia plays a role similar to that of mass in linear motion. It measures how much an object resists changes to its rotational motion. For a uniform rod rotating about one of its ends, the moment of inertia is determined using the formula:- \[ I = \frac{1}{3} m L^2 \]
Where:
  • \(m\) is the mass of the rod.
  • \(L\) is the length of the rod.
In this problem, the bat is approximated as a uniform rod with a mass of 2.2 kg and a length of 0.95 m. This approximation allows us to plug these values into the formula. A higher moment of inertia means that more torque is needed to change its rotational speed. Calculating this important parameter is the first step to understanding the rotational dynamics of the bat.
Angular Acceleration
Angular acceleration describes how quickly the rotational speed of an object is changing. It is similar to linear acceleration but applied to rotational motion. To find angular acceleration, we use the change in angular velocity over time:- \[ \alpha = \frac{\omega_f - \omega_i}{t} \]
Where:
  • \(\omega_f\) is the final angular velocity.
  • \(\omega_i\) is the initial angular velocity (often zero if it starts from rest).
  • \(t\) is the time over which the change occurs.
In this exercise, the softball bat starts from rest and reaches a final rotational speed of 2.7 revolutions per second, which we convert to radians per second. With these values, we determine how fast the rotational velocity is increasing, which is crucial for calculating torque.
Uniform Rod
A uniform rod is a simple model used to simplify complex real-world objects like a baseball bat. By assuming the mass is distributed evenly throughout its length, calculations like the moment of inertia become straightforward. This simplification allows us to apply standard physics formulas effectively.

For example, when we model a bat as a uniform rod, we assume it has consistent properties that make it easier to analyze. Although an actual bat may not be perfectly uniform, this approximation is useful for theoretical calculations and provides a good estimate of the physical behavior.

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Most popular questions from this chapter

A small 650 -g ball on the end of a thin, light rod is rotated in a horizontal circle of radius \(1.2 \mathrm{~m}\). Calculate \((a)\) the moment of inertia of the ball about the center of the circle, and \((b)\) the torque needed to keep the ball rotating at constant angular velocity if air resistance exerts a force of \(0.020 \mathrm{~N}\) on the ball. Ignore the rod's moment of inertia and air resistance.

(II) A thin, hollow 0.545 -kg section of pipe of radius 10.0 \(\mathrm{cm}\) starts rolling (from rest) down a \(17.5^{\circ}\) incline 5.60 \(\mathrm{m}\) long. (a) If the pipe rolls without slipping, what will be its speed at the base of the incline? (b) What will be its total kinetic energy at the base of the incline? (c) What minimum value must the coefficient of static friction have if the pipe is not to slip?

A narrow but solid spool of thread has radius \(R\) and mass \(M\). If you pull up on the thread so that the \(\mathrm{CM}\) of the spool remains suspended in the air at the same place as it unwinds, (a) what force must you exert on the thread? (b) How much work have you done by the time the spool turns with angular velocity \(\omega ?\)

(a) A grinding wheel \(0.35 \mathrm{~m}\) in diameter rotates at 2500 rpm. Calculate its angular velocity in rad/s. (b) What are the linear speed and acceleration of a point on the edge of the grinding wheel?

(III) The 1100 -kg mass of a car includes four tires, each of mass (including wheels) 35 \(\mathrm{kg}\) and diameter 0.80 m. Assume each tire and wheel combination acts as a solid cylinder. Determine \((a)\) the total kinetic energy of the car when traveling 95 \(\mathrm{km} / \mathrm{h}\) and \((b)\) the fraction of the kinetic energy in the tires and wheels. (c) If the car is initially at rest and is then pulled by a tow truck with a force of \(1500 \mathrm{N},\) what is the acceleration of the car? Ignore frictional losses. \((d)\) What percent error would you make in part \((c)\) if you ignored the rotational inertia of the tires and wheels?

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