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<h2 class="hd hd-2 unit-title">Intro to Circular Motion - Acceleration </h2>
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<p> As you go through the lesson, you will find this list of reading from the textbook complementary to the videos and exercises, and on many occasions these readings will contain relevant material to the presentation:</p><ul><li> Tangential and Radial Acceleration: <a href="/courses/course-v1:MITx+8.01.1x+3T2018/pdfbook/0/chapter/6/6"><i class="it">chapter 6, section 3</i></a></li><li> Angular Velocity and Angular Acceleration: <a href="/courses/course-v1:MITx+8.01.1x+3T2018/pdfbook/0/chapter/6/11"><i class="it">chapter 6, section 5</i></a></li></ul>
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<h2 class="hd hd-2 unit-title">L10v1: Circular Motion - Acceleration</h2>
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<p>Acceleration in Circular Motion: \(\vec{a} = r\frac{d^2\theta}{dt^2}\hat{\theta} + r(\frac{d\theta}{dt})^2(-\hat{r})\)</p>
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<h3 class="hd hd-2">L10v1: Circular Motion - Acceleration</h3>
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<p><b>For circular motion, show that \( \frac{d \hat{\theta}}{dt} = -\frac{d \theta}{dt}\hat{r}\) </b></p><p>
We know that \( \hat{\theta } \) changes direction as we go around a circle, and so if a particle is undergoing circular motion, we expect \(\frac{d\hat{\theta }}{dt}\) to be non-zero. To figure out exactly what it is, let us write \(\hat{\theta }\) in terms of \(\hat{i}\) and \(\hat{j}\) for an arbitrary \(\theta\). </p><center><img src="/assets/courseware/v1/fcb4ed339b51092ff8c740bedcd8eca0/asset-v1:MITx+8.01.1x+3T2018+type@asset+block/images_text_polarCoords1.svg" width="440"/></center><p> \( \begin{align} \hat{\theta} = -\sin(\theta) \hat{i} + \cos(\theta) \hat{j} \end{align} \) </p><p> Since \(\hat{i}\) and \(\hat{j}\) are constant anywhere in space, they are not time-dependent, so now the derivative becomes much more clear. Using the chain rule, we have: </p><p> \( \begin{align} \frac{d\hat{\theta}}{dt} = -\cos(\theta) \frac{d\theta}{dt} \hat{i} - \sin(\theta) \frac{d\theta}{dt} \hat{j} = - \frac{d\theta}{dt} \left(\cos(\theta) \hat{i} + \sin(\theta)\hat{j} \right) \end{align} \) </p><p> We know that \( \left(\cos(\theta) \hat{i} + \sin(\theta)\hat{j} \right) = \hat{r} \), so we can now simply write this derivative as: </p><p> \[ \begin{align} \frac{d\hat{\theta}}{dt} = - \frac{d\theta}{dt} \hat{r} \end{align} \]</p>
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<h2 class="hd hd-2 unit-title">L10Q1: Direction of the Acceleration</h2>
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L10Q1: Direction of the Acceleration
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An object moves counter-clockwise along the circular path as shown. As it moves along the path its acceleration vector continuously points toward point D. The object </p>
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<text> speeds up at B.</text>
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<text> slows down at B.</text>
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<text> No object can execute such a motion.</text>
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<h2 class="hd hd-2 unit-title">L10Q2: Radial and Tangential Components</h2>
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L10Q2: Tangential and Radial Components
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A particle is moving in a circle of radius [mathjaxinline]R[/mathjaxinline]. At [mathjaxinline]t = 0[/mathjaxinline], it is located on the [mathjaxinline]+x[/mathjaxinline]-axis. The angle the particle makes with the [mathjaxinline]+x[/mathjaxinline]-axis is given by [mathjaxinline]\theta (t) = At^2[/mathjaxinline], where [mathjaxinline]A[/mathjaxinline] is a positive constant. </p>
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Its acceleration vector has a tangential and radial component, [mathjaxinline]a_{\theta }[/mathjaxinline] and [mathjaxinline]a_ r[/mathjaxinline] which can be written as: </p>
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[mathjaxinline]\displaystyle \vec{a} = a_{\theta }\hat{\theta }+a_ r\hat{r}[/mathjaxinline]
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Which of the following statements best describes the tangential and the radial components of the acceleration at the instant [mathjaxinline]t = 1[/mathjaxinline] sec. </p>
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<text> [mathjaxinline]a_{\theta }=0[/mathjaxinline], [mathjaxinline]a_ r = 0[/mathjaxinline]</text>
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<text> [mathjaxinline]a_{\theta }&gt;0[/mathjaxinline], [mathjaxinline]a_ r &lt; 0[/mathjaxinline]</text>
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<text> [mathjaxinline]a_{\theta }&lt;0[/mathjaxinline], [mathjaxinline]a_ r &lt; 0[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L10v2: Angular Acceleration</h2>
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<h2 class="hd hd-2 unit-title">L10Q3: Angular Acceleration</h2>
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L10Q3: Angular Acceleration
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A disc is rotating about an axis parallel to the z-axis and passing through its center. </p>
<p><b class="bfseries">(Part a)</b> In which of the figures above is the disc slowing down and rotating clockwise as viewed from the [mathjaxinline]+z[/mathjaxinline]-axis <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_ls_ls03_ls03_09_2_1">
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<p><b class="bfseries">(Part b)</b> In which of the figures above is the disc speeding up and rotating clockwise as viewed from the [mathjaxinline]+z[/mathjaxinline]-axis? </p>
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<text> Figure a)</text>
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<h2 class="hd hd-2 unit-title">L10WE1/L10v3: Worked Example - Angular Position from Angular Acceleration (Worked Example)</h2>
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L10WE1: Angular Position from angular acceleration (Worked Example)
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A particle is moving in a circle of radius [mathjaxinline]r[/mathjaxinline] with an angular acceleration given by [mathjaxinline]\vec{\alpha }(t)=\dfrac {1}{r}(A-Bt)\hat{k}[/mathjaxinline], or [mathjaxinline]a_{\theta }=(A-Bt)\hat{\theta }[/mathjaxinline] where [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]B[/mathjaxinline] are positive constants. At time [mathjaxinline]t_0=0[/mathjaxinline], the particle is at an angle [mathjaxinline]\theta (0)[/mathjaxinline] measured with respect to the [mathjaxinline]+x[/mathjaxinline]-axis, and the tangential component of its velocity is [mathjaxinline]v_{\theta }(0)=v_0[/mathjaxinline]. </p>
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Find the particle's angular position [mathjaxinline]\theta (t)[/mathjaxinline] as a function of time to obtain the arc length travelled by the particle during the time [mathjaxinline]t[/mathjaxinline] and defined as [mathjaxinline]s(t)=r(\theta (t)-\theta (0))[/mathjaxinline]. Express your answer in terms of [mathjaxinline]r[/mathjaxinline], [mathjaxinline]A[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]t[/mathjaxinline], and v_0 for [mathjaxinline]v_0[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]s(t) =[/mathjaxinline] </p>
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<h3 class="hd hd-2">L10v3: Worked Example - Angular position from angular acceleration.</h3>
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<h2 class="hd hd-2 unit-title">L10Q4: Angular Position from Angular Acceleration</h2>
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L10Q4: Angular Position from Angular Acceleration
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A particle is moving in a circle of radius [mathjaxinline]r[/mathjaxinline] contained in the [mathjaxinline]xy[/mathjaxinline]-plane. At the initial time, [mathjaxinline]t_0 = 0[/mathjaxinline], the particle is at rest on the [mathjaxinline]x[/mathjaxinline]-axis, [mathjaxinline]\theta (0) = 0[/mathjaxinline]. The particle's angular acceleration vector for [mathjaxinline]t\ge 0[/mathjaxinline] is given by: [mathjaxinline]\vec{\alpha }(t)=(A-Bt)\hat{k}[/mathjaxinline], where [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]B[/mathjaxinline] are positive constants. </p>
<p><b class="bfseries">(Part a)</b> Drag and drop the proper units of the constants [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]B[/mathjaxinline]: </p>
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<p><b class="bfseries">(Part b)</b> Which of the following statements about the particle's motion as viewed from the +z-axis is true: <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_ls_ls03_ls03_16_3_1">
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<input type="radio" name="input_ls_ls03_ls03_16_3_1" id="input_ls_ls03_ls03_16_3_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="ls_ls03_ls03_16_3_1-choice_2-label" for="input_ls_ls03_ls03_16_3_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_ls_ls03_ls03_16_3_1"> <text> immediately after [mathjaxinline]t=0[/mathjaxinline] it moves counterclockwise, it stops momentarily, then it moves clockwise</text>
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<input type="radio" name="input_ls_ls03_ls03_16_3_1" id="input_ls_ls03_ls03_16_3_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="ls_ls03_ls03_16_3_1-choice_3-label" for="input_ls_ls03_ls03_16_3_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_ls_ls03_ls03_16_3_1"> <text> it moves counterclockwise for all [mathjaxinline]t \gt 0[/mathjaxinline]</text>
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<input type="radio" name="input_ls_ls03_ls03_16_3_1" id="input_ls_ls03_ls03_16_3_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="ls_ls03_ls03_16_3_1-choice_4-label" for="input_ls_ls03_ls03_16_3_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_ls_ls03_ls03_16_3_1"> <text> it moves clockwise for all [mathjaxinline]t \gt 0[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L10Q5: Moving around a Circle</h2>
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L10Q5: Moving around a Circle
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The particle in the movie above starts from rest at [mathjaxinline]t = 0[/mathjaxinline], it moves counterclockwise along the circle, and at time [mathjaxinline]t=t_1[/mathjaxinline] it stops momentarily. For times [mathjaxinline]t&gt;t_1[/mathjaxinline], the particle speeds up moving clockwise. </p>
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Consider the particle's acceleration to be [mathjaxinline]\vec{a} = a_ r\hat{r}+a_\theta \hat{\theta }[/mathjaxinline], its angular acceleration [mathjaxinline]\vec{\alpha }=\alpha _ z\hat{k}[/mathjaxinline], and its angular velocity [mathjaxinline]\vec{\omega }=\omega _ z\hat{k}[/mathjaxinline]. Drag and drop the corresponding signs of the components of the vectorial quantities [mathjaxinline]\vec{a}[/mathjaxinline], [mathjaxinline]\vec{\alpha }[/mathjaxinline] and [mathjaxinline]\vec{\omega }[/mathjaxinline] at the instant when the particle passes through point B for the first time ([mathjaxinline]t&lt;t_1[/mathjaxinline]), and at the instant when it passes through point B for the second time ([mathjaxinline]t&gt;t_1[/mathjaxinline]). Use zero if the component is zero. </p>
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<h2 class="hd hd-2 unit-title">L10: Review of Rotational Kinematics</h2>
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<p> Review of Rotational Kinematics.</p><p>Position: \(\vec{r}(t)=r\hat{r}\)</p><p> Angular Velocity: \(\vec{\omega}=\omega_z \hat{k}\)</p><p>Component of the Angular Velocity along the axis of rotation: \(\omega_z=\dfrac{d\theta}{dt} \)</p><p>Velocity: \(\vec{v}=r\dfrac{d\theta}{dt}\hat{\theta} \)</p><p> Angular Acceleration: \(\vec{\alpha} = \alpha_{z}\hat{k} \)</p><p>Component of the Angular Acceleration: \(\alpha_z=\dfrac{d\omega_z}{dt}=\dfrac{d^2\theta}{dt^2}\)</p><p>Acceleration: \(\vec{a}=a_r\hat{r}+a_{\theta}\hat{\theta} \)</p><p> \(a_r = -\frac{v^2}{r} = -r(\dfrac{d\theta}{dt})^2, a_{\theta} = r\dfrac{d^2\theta}{dt^2}= r\alpha_z\)
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