Circular Motion: 17 flashcards
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1. What is centripetal acceleration?
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The acceleration of an object moving in a circular path, directed toward the center of the circle. Its magnitude is $a_c = \frac{v^2}{r}$.
2. What does the symbol $v$ represent in the centripetal acceleration formula?
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The magnitude of the constant speed of the object along the circular path.
3. What does the symbol $r$ represent in the centripetal acceleration formula?
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The radius of the circular path.
4. In the derivation, why is the velocity vector always perpendicular to the radius?
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Because for uniform circular motion, the velocity is tangent to the circle at every point, and the radius is radial; they are perpendicular by geometry.
5. What is the key geometric relationship used to derive the centripetal acceleration formula?
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The triangle formed by the velocity vectors and the triangle formed by the radii are similar (isosceles triangles with equal angles).
6. In the derivation, what does $\Delta v$ represent?
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The change in velocity vector over a small time interval $\Delta t$; its magnitude is $v \Delta \theta$.
7. How is the angle $\Delta \theta$ related to the arc length $\Delta s$?
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$\Delta \theta = \frac{\Delta s}{r}$ (in radians).
8. What is the direction of the acceleration vector in uniform circular motion?
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It points toward the center of the circle, perpendicular to the velocity vector.
9. Write the formula for centripetal acceleration in terms of angular speed $\omega$.
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$a_c = \omega^2 r$
10. In the derivation, why is $\Delta v$ approximately equal to $v \Delta \theta$ for small $\Delta \theta$?
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Because for small angles, the chord length (magnitude of $\Delta \vec{v}$) approximates the arc length, and $\Delta \theta$ is small, so $|\Delta \vec{v}| \approx v \Delta \theta$.
11. What is the average acceleration during a small time interval $\Delta t$ in the derivation?
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$\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t}$
12. How do you obtain the instantaneous centripetal acceleration from the average acceleration?
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Take the limit as $\Delta t \to 0$: $\vec{a} = \lim_{\Delta t \to 0} \frac{\Delta \vec{v}}{\Delta t}$.
13. In the derivation, what is the relationship between $\Delta t$ and $\Delta \theta$?
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$\Delta \theta = \omega \Delta t$, where $\omega$ is the angular speed.
14. Show the step-by-step derivation that leads to $a_c = \frac{v^2}{r}$.
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1. $|\Delta \vec{v}| \approx v \Delta \theta$ (small angle). 2. $\Delta \theta = \frac{\Delta s}{r} = \frac{v \Delta t}{r}$. 3. So $|\Delta \vec{v}| \approx \frac{v^2}{r} \Delta t$. 4. Then $a_c = \lim_{\Delta t \to 0} \frac{|\Delta \vec{v}|}{\Delta t} = \frac{v^2}{r}$.
15. If the speed of an object in circular motion doubles, what happens to the centripetal acceleration?
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It becomes four times larger, because $a_c \propto v^2$.
16. If the radius of the circular path is halved while keeping the speed constant, what happens to the centripetal acceleration?
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It doubles, because $a_c \propto \frac{1}{r}$.
17. Why is the centripetal acceleration not zero even if the speed is constant?
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Because acceleration is a vector quantity; the direction of the velocity changes continuously, so there is a nonzero acceleration even though the magnitude of velocity (speed) remains constant.
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