Topic: Dynamics
Concepts: Circular motion, Conservation of energy, Newton’s laws

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Solution:

The pendulum feels the forces of tension and gravity. Applying Newton’s 2nd law in the tangential direction,

Since the pendulum is undergoing circular motion, the radial acceleration is given by

The magnitude of the total acceleration is

Now we go through the possible answer choices.

• A) Not correct since the magnitude of the acceleration depends on $\theta$ and $v$ which are both changing.
• B) Not correct since the magnitude of the acceleration at the lowest point $\theta=0$ is
  
  $a=v^2/L$
  
  and we have from energy conservation,
  
  $\frac{1}{2}mv^2=mgL(1-\cos\theta_{\text{max}})$
  $v^2/L=2g(1-\cos\theta_{\text{max}})$
  
  which is not (in general) equal to $g$.
• C) Not correct since $v=0$ (no radial acceleration) only at $\theta=\theta_{\text{max}}$ (where there is a tangential acceleration) so $a \neq 0$ at any point.
• D) Not correct since the mass has a tangential acceleration $a_t=g\sin\theta$.
• E) Correct since at the lowest point $\theta=0$ there is no tangential acceleration $a_t=0$.

Thus, the answer is E.

Topic: Dynamics
Concepts: Circular motion

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Solution:

The pendulum is undergoing circular motion, so the radial acceleration $\vec{a}_r$ points toward the center of the circle. The pendulum is slowing down as it moves higher so the tangential acceleration $\vec{a}_t$ points opposite the velocity. The total acceleration

is the vector sum of these two components. Thus, the answer is D.