Question 1.1a A large grindstone rotates at 80 revolutions per minute. You hold a knife against the stone to steadily slow it down until it stops in \( 20 \mathrm{~s} \). a) What is the initial angular velocity of the stone in \( \mathrm{rad} / \mathrm{s} \) ? E 2. 1.1b What is the angular acceleration of the stone in \( \mathrm{rad} / \mathrm{s}^{2} \) ? 3. 1.1c How many revolutions does the stone go through during this process? 1.2a The Earth's radius is \( 6371 \mathrm{~km} \) and its mass is \( 5.97 \times 10^{24} \mathrm{~kg} \). The following questions are about the Earth's rotaiton around its own axis, NOT its orbit around the sun. Assume the Earth is spherical. What is the Earth's angular velocity in \( \mathrm{rad} / \mathrm{s} \) ? 5. 1.2b What is the Earth's moment of inertia about its axis of rotation in \( \mathrm{kg} * \mathrm{~m}^{2} \) ? 6. 1.2c What is the Earth's rotational kinetic energy in \( \mathrm{J} \) ? 7. 1.2d What is the speed of Rice University as it circles with the Earth in \( \mathrm{m} / \mathrm{s} \) ? Rice is at a latitude of 29.7 degrees

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Transcribed Image Text: 1.1a A large grindstone rotates at 80 revolutions per minute. You hold a knife against the stone to steadily slow it down until it stops in \( 20 \mathrm{~s} \). a) What is the initial angular velocity of the stone in \( \mathrm{rad} / \mathrm{s} \) ? E 2. 1.1b What is the angular acceleration of the stone in \( \mathrm{rad} / \mathrm{s}^{2} \) ? 3. 1.1c How many revolutions does the stone go through during this process? 1.2a The Earth's radius is \( 6371 \mathrm{~km} \) and its mass is \( 5.97 \times 10^{24} \mathrm{~kg} \). The following questions are about the Earth's rotaiton around its own axis, NOT its orbit around the sun. Assume the Earth is spherical. What is the Earth's angular velocity in \( \mathrm{rad} / \mathrm{s} \) ? 5. 1.2b What is the Earth's moment of inertia about its axis of rotation in \( \mathrm{kg} * \mathrm{~m}^{2} \) ? 6. 1.2c What is the Earth's rotational kinetic energy in \( \mathrm{J} \) ? 7. 1.2d What is the speed of Rice University as it circles with the Earth in \( \mathrm{m} / \mathrm{s} \) ? Rice is at a latitude of 29.7 degrees

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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/5IntroductionAngular velocity is a measure of the rate of change of angular displacement, which is the angle through which an object rotates or revolves around a fixed axis. It is a vector quantity that describes the speed and direction of rotation of an object, and is measured in radians per second (rad/s) or degrees per second (°/s). Angular velocity is calculated by dividing the change in angular displacement by the time taken for the change to occur, and is used in a variety of applications, including physics, engineering, and astronomy.Angular velocity is the rate of change of angular displacement with respect to time. It is a measure of how fast an object is rotating around a fixed axis. Angular velocity is a vector quantity, meaning that it has both magnitude and direction. It is measured in radians per second (rad/s) or degrees per second (°/s). The formula for angular velocity is:\( \begin{align*} \mathrm{\omega} &= \mathrm{\Delta\theta\div\Delta{t}} \end{align*} \)where ω is the angular velocity, Δθ is the change in angular displacement, and Δt is the time taken for the change to occur. The direction of the angular velocityis perpendicular to the plane of rotation and follows the right-hand rule. Angular velocity is used in many fields, including physics, engineering, and astronomy.Explanationhe direction of the angular velocityis perpendicular to the plane of rotation and follows the right-hand rule. Angular velocity is used in many fields, including physics, engineering, and astronomy.Explanation:Please refer to solution in this step.Step2/5 a)To find the initial angular velocity of the stone, we can use the formula:\( \begin{align*} \mathrm{\omega} &= \mathrm{\Delta\theta\div\Delta{t}} \end{align*} \)where ω is the angular velocity in radians per second, Δθ is the change in angle (in radians), and Δt is the time interval.At the start of the process, the grindstone is rotating at 80 revolutions per minute, which is equivalent to:\( \begin{align*} \mathrm{\omega{\left({i}\right)}} &= \mathrm{{\left({80}{r}{e}{v}\div\min\right)}{x}{\left({2}\Pi{r}{a}{d}\div{r}{e}{v}\right)}\div{\left({60}{s}\right)}} \end{align*} \)\( \begin{align*} \mathrm{\omega{\left({i}\right)}} &= \mathrm{{8.3776}{r}{a}{d}{p}{e}{r}{\sec{{o}}}{n}{d}} \end{align*} \)During the 20 seconds it takes for the stone to stop, the angle it rotates through is:\( \begin{align*} \mathrm{\Delta\theta} &= \mathrm{\omega{\left({i}\right)}{x}\Delta{t}={8.3776}{x}{20}} \end{align*} \)\( \begin{align*} \mathrm{\Delta\theta} &= \mathrm{{167.552}{r}{a}{d}} \end{align*} \)Explanationwe got the results for change in angle \( \begin{align*} \mathrm{\Delta\theta} &= \mathrm{{167.552}} \end{align*} \)Explanation:Please refer to solution in this step.Step3/5b) To find the angular acceleration of the stone, we can use the formula:\( \begin{align*} \mathrm{\alpha} &= \mathrm{\Delta\omega\div\Delta{t}} \end{align*} \)where α is the angular acceleration in radians per second squared, and Δω is the change in angular velocity.At the end of the process, the stone has stopped rotating, so its final angular velocity is zero. Therefore:\( \begin{align*} \mathrm{\Delta\omega} &= \mathrm{{0}-\omega{\left({i}\right)}} \end{align*} \)\( \begin{align*} \mathrm{\Delta\omega} &= \mathrm{-{8.3776}{r}{a}{d}{p}{e}{r}{\sec{{o}}}{n}{d}} \end{align*} \)The time interval is still 20 seconds, so:\( \begin{align*} \mathrm{\alpha} &= \mathrm{{\left(-{8.37 ... See the full answer