Ch 04: Kinematics in Two Dimensions

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 04: Kinematics in Two Dimensions

Problem 75

Chapter 4, Problem 75

A painted tooth on a spinning gear has angular position θ = (6.0 rad/s⁴)t⁴. What is the tooth's angular acceleration at the end of 10 revolutions?

Verified step by step guidance

Step 1: Understand the given angular position equation θ = (6.0 rad/s⁴)t⁴. Here, θ is the angular position as a function of time t, and the coefficient 6.0 rad/s⁴ represents the proportionality constant. Angular acceleration is the second derivative of angular position with respect to time.

Step 2: Differentiate θ with respect to time t to find the angular velocity ω. Using the formula ω = dθ/dt, calculate the first derivative of θ = (6.0 rad/s⁴)t⁴. This gives ω = 4(6.0 rad/s⁴)t³ = 24.0t³ rad/s.

Step 3: Differentiate ω with respect to time t to find the angular acceleration α. Using the formula α = dω/dt, calculate the second derivative of θ. This gives α = d(24.0t³)/dt = 72.0t² rad/s².

Step 4: Determine the time t corresponding to 10 revolutions. One revolution corresponds to an angular displacement of 2π radians. Therefore, 10 revolutions correspond to an angular displacement of θ = 10 × 2π = 20π radians. Solve for t in the equation θ = (6.0 rad/s⁴)t⁴ by substituting θ = 20π. This gives 20π = (6.0 rad/s⁴)t⁴. Solve for t.

Step 5: Substitute the value of t obtained in Step 4 into the angular acceleration formula α = 72.0t² rad/s² to find the angular acceleration at the end of 10 revolutions.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Angular Position

Angular position describes the orientation of a rotating object in terms of an angle, typically measured in radians. In this question, the angular position θ is given as a function of time, indicating how it changes as the gear spins. Understanding this concept is crucial for determining how the gear's rotation evolves over time.

Angular Velocity

Angular velocity is the rate of change of angular position with respect to time, usually expressed in radians per second (rad/s). It can be derived from the angular position function by taking its first derivative. In this case, knowing the angular velocity at a specific time will help in calculating the angular acceleration.

Angular Acceleration

Angular acceleration is the rate of change of angular velocity with respect to time, measured in radians per second squared (rad/s²). It can be found by taking the derivative of the angular velocity. In this problem, calculating the angular acceleration at the end of 10 revolutions requires understanding how the angular position function relates to both angular velocity and acceleration.

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Textbook Question

In Problems 78, 79, and 80 you are given the equations that are used to solve a problem. For each of these, you are to write a realistic problem for which these are the correct equations. Be sure that the answer your problem requests is consistent with the equations given.

$\begin{aligned}\)100 \, $\text{m}$ &= 0 \, $\text{m}$ + (50 $\cos$ $\theta$ \, $\text{m/s}$) t_1 \\0 \, $\text{m}$ &= 0 \, $\text{m}$ + (50 $\sin$ $\theta$ \, $\text{m/s}$) t_1 - $\frac{1}{2}$ (9.80 \, $\text{m/s}$^2) t_1^2\(\end{aligned}$

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Textbook Question

A 6.0-cm-diameter gear rotates with angular velocity ω = ( 20 ─ ½ t² ) rad/s where t is in seconds. At t = 4.0 s, what are: The tangential acceleration of a tooth on the gear?

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Textbook Question

A 6.0-cm-diameter gear rotates with angular velocity ω = ( 20 ─ ½ t² ) rad/s where t is in seconds. At t = 4.0 s, what are: The gear's angular acceleration?

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