Day: January 12, 2023

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202301120908 Solution to 2002-IBHL-PHY-I-9

A pendulum has a bob of mass m and swings in the arc shown below.

As the bob swings through the lowest point of its motion, the tension in the string will be

A. zero.
B. less than mg.
C. equal to mg.
D. greater than mg.

Roughwork.

Assumed the string is l long, taut and inextensible.

Derivation. Read More

    \begin{aligned} T & = \frac{1}{2}ml^2\dot{\theta}^2 \\ V & = mgl(1-\cos\theta ) \\ \mathcal{L} & = T-V \\ \frac{\partial \mathcal{L}}{\partial\theta} & = -mgl\sin\theta \\ \frac{\partial \mathcal{L}}{\partial\dot{\theta}} & = ml^2\dot{\theta} \\ 0 & =\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial \dot{\theta}}\bigg) - \frac{\partial\mathcal{L}}{\partial \theta} \\ \end{aligned}

    A simple pendulum, under gravity, has

    \textrm{EoM: }\qquad\displaystyle{\ddot{\theta}+\frac{g}{l}\cdot\sin\theta =0}

    the equation of motion, independent of mass m, neglecting friction.

    Lemma. (not in use) Read More

    \begin{aligned} \begin{bmatrix} \hat{\mathbf{e}}_r \\ \hat{\mathbf{e}}_\theta \end{bmatrix} & = \begin{bmatrix} \displaystyle{\frac{\partial x}{\partial r}} & \displaystyle{\frac{\partial y}{\partial r}} \\[2mm]\displaystyle{\frac{1}{r}\frac{\partial x}{\partial\theta}} & \displaystyle{\frac{1}{r}\frac{\partial y}{\partial\theta}} \end{bmatrix}\begin{bmatrix} \hat{\mathbf{e}}_x \\ \hat{\mathbf{e}}_y \end{bmatrix} \\ \begin{bmatrix} \hat{\mathbf{e}}_x \\ \hat{\mathbf{e}}_y \end{bmatrix} & = \begin{bmatrix} \displaystyle{\frac{1}{r}\frac{\partial y}{\partial \theta}} & \displaystyle{-\frac{\partial y}{\partial r}} \\[2mm]\displaystyle{-\frac{1}{r}\frac{\partial x}{\partial\theta}} & \displaystyle{\frac{\partial x}{\partial r}} \end{bmatrix} \begin{bmatrix} \hat{\mathbf{e}}_r \\ \hat{\mathbf{e}}_\theta \end{bmatrix} \\ \end{aligned}

    The net force for this non-uniform circular motion, being centripetal force producing radial plus tangential accelerations,

    \mathbf{a} = -l(\dot{\theta}^2\cos\theta +\ddot{\theta}\sin\theta)\,\hat{\mathbf{i}} + l(\ddot{\theta}\cos\theta -\dot{\theta}^2\sin\theta )\,\hat{\mathbf{j}}

    is the resultant of tension

    \begin{aligned} T & = |\mathbf{T}| \\ \mathbf{T} & =T\sin\theta\,\hat{\mathbf{i}}+T\cos\theta\,\hat{\mathbf{j}}} \end{aligned}

    and weight \mathbf{W}=m\mathbf{g}=-mg\,\hat{\mathbf{j}}. Write, with a free-body diagram in mind,

    \begin{aligned} \because\enspace & \mathbf{F}_{\textrm{net}} =\mathbf{T}+\mathbf{W}=m\mathbf{a} \\ \therefore\enspace & \begin{cases} T\sin\theta = -ml(\dot{\theta}^2\cos\theta+\ddot{\theta}\sin\theta ) \\ T\cos\theta - mg = ml(\ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta ) \\ \end{cases} \\ \end{aligned}

    When the bob is at the lowest point, i.e., \theta =0,

    \begin{aligned} T & = mg + ml\ddot{\theta} \\ & = mg - mg\sin\theta \qquad (\textrm{by EoM})\\ & \approx mg-mg\theta \qquad (\because\enspace \sin\theta\approx\theta\textrm{ for small }\theta)\\ & \lneq mg \\ \end{aligned}

    The official answer is D but mine B.  Could that be E: more or less mg? If I am wrong, I must have omitted some simple facts of noble truth.

    This problem is not to be attempted.