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AP Physics C: Mechanics : Understanding and Finding Center of Mass

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2 Diagnostic Tests 92 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Circular And Rotational Motion

Doing which of the following would allow you to find the center of mass of an object?

Possible Answers:

Recording its shape

Sliding it along a flat surface

Spinning it

Hanging it from a fixed point

Correct answer:

Spinning it

Explanation:

Center of mass can be found by spinning an object. It will naturally spin around its center of mass, due to the concept of even distribution of mass in relation to the center of mass. Shape and mass are important factors in this property, but the most improtant factor is the mass distribution.

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Example Question #1 : Rotational Motion And Torque

If the fulcrum of a balanced scale is shifted to the left, what type of adjustment must be made to rebalance the scale?

Possible Answers:

Apply more mass to the right end

Apply the same amount of mass to both ends

Apply more mass to the new position of the fulcrum

Apply more mass to the left end

If the scale was initially balanced, moving the fulcrum will not change this

Correct answer:

Apply more mass to the left end

Explanation:

Changing the position of the fulcrum by moving it to the left means the center of mass will be to the right of the new position. Therefore, the scale will tip right. Adding more mass to the left end will rebalance the scale. None of the other options make sense. Adding more mass to the new fulcrum position will not change the balance of the scale because that mass is a negligible distance from the new fulcrum position and does nothing to change the masses on either side.

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Example Question #54 : Ap Physics C

If two masses, $m_{1}=50kg$ $\displaystyle m_{1}=50kg$ and $m_{2}=200kg$ $\displaystyle m_{2}=200kg$ are placed on a seesaw of length $\displaystyle L$, where must the fulcrum be placed such that the seesaw remains level?

Possible Answers:

$\frac{5}{7}L$ $\displaystyle \frac{5}{7}L$

$\frac{4}{5}L$ $\displaystyle \frac{4}{5}L$

$\frac{2}{3}L$ $\displaystyle \frac{2}{3}L$

$\frac{L}{5}$ $\displaystyle \frac{L}{5}$

$\frac{L}{4}$ $\displaystyle \frac{L}{4}$

Correct answer:

$\frac{4}{5}L$ $\displaystyle \frac{4}{5}L$

Explanation:

This question asks us to find the center of mass for this system. We know that the center of mass resides a distance $x_{1}$ $\displaystyle x_{1}$ from the first mass such that:

$m_{1}x_{1}=m_{2}(L-x_{1})$ $\displaystyle m_{1}x_{1}=m_{2}(L-x_{1})$

In this case:

$m_{1}x_{1}+m_{2}x_{1}=m_{2}L$ $\displaystyle m_{1}x_{1}+m_{2}x_{1}=m_{2}L$

$x_{1}(m_{1}+m_{2})=m_{2}L$ $\displaystyle x_{1}(m_{1}+m_{2})=m_{2}L$

Plug in known values and solve.

$x_{1}=\frac{m_{2}L}{m_{1}+m_{2}}=\frac{200}{250}L=\frac{4}{5}L$ $\displaystyle x_{1}=\frac{m_{2}L}{m_{1}+m_{2}}=\frac{200}{250}L=\frac{4}{5}L$

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Example Question #55 : Ap Physics C

Three $40\: g$ $\displaystyle 40\: g$ point masses are at the points $(7.07 \:cm, 0\: cm)$ $\displaystyle (7.07 \:cm, 0\: cm)$, $(-7.07 \:cm, 0 \:cm)$ $\displaystyle (-7.07 \:cm, 0 \:cm)$ and

$(0 \:cm, -7.07 \:cm)$ $\displaystyle (0 \:cm, -7.07 \:cm)$and a $90 \:g$ $\displaystyle 90 \:g$ point mass is at the point $(0 \:cm, 7.07 \:cm)$ $\displaystyle (0 \:cm, 7.07 \:cm)$.

How far from the origin is the center of mass of the system?

Possible Answers:

$0.786\:cm$ $\displaystyle 0.786\:cm$

$1.41\:cm$ $\displaystyle 1.41\:cm$

$1.68\:cm$ $\displaystyle 1.68\:cm$

$1.12\:cm$ $\displaystyle 1.12\:cm$

$0.909\:cm$ $\displaystyle 0.909\:cm$

Correct answer:

$1.68\:cm$ $\displaystyle 1.68\:cm$

Explanation:

To find the center of mass, we have to take the weighted average of the x coordinates and the y coordinates.

$40\:g$ $\displaystyle 40\:g$ Measures: $90\:g$ $\displaystyle 90\:g$ Measures

$(7.07 \:cm, 0\: cm)$ $\displaystyle (7.07 \:cm, 0\: cm)$ $(0 \:cm, 7.07 \:cm)$ $\displaystyle (0 \:cm, 7.07 \:cm)$

$(-7.07 \:cm, 0 \:cm)$ $\displaystyle (-7.07 \:cm, 0 \:cm)$

$(0 \:cm, -7.07 \:cm)$ $\displaystyle (0 \:cm, -7.07 \:cm)$

First, we take the weighted measurement of the x-axis:

$x_c_m = \frac{(7.07\cdot40\:g)+(-7.07 \cdot 40\:g) + (0\cdot 40\:g)+(0\cdot90\:g)}{(3\cdot40)+90}$ $\displaystyle x_c_m = \frac{(7.07\cdot40\:g)+(-7.07 \cdot 40\:g) + (0\cdot 40\:g)+(0\cdot90\:g)}{(3\cdot40)+90}$

We can see that the result of the x-axis contribution is equal to $\displaystyle 0$.

Now, let's look at the y-axis contribution:

$y_c_m = \frac{(0\cdot40\:g)+(0 \cdot 40\:g) + (-7.07\cdot 40\:g)+(7.07\cdot90\:g)}{(3\cdot40)+90}$ $\displaystyle y_c_m = \frac{(0\cdot40\:g)+(0 \cdot 40\:g) + (-7.07\cdot 40\:g)+(7.07\cdot90\:g)}{(3\cdot40)+90}$

This equals to $1.68\:cm$ $\displaystyle 1.68\:cm$

Now that we have the x and y components, we take the root of squares to get the final answer:

$\uptext{center}= \sqrt{x^2+y^2}$ $\displaystyle \uptext{center}= \sqrt{x^2+y^2}$

This will give us $1.68\:cm$ $\displaystyle 1.68\:cm$

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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Understanding and Finding Center of Mass

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #2 : Circular And Rotational Motion

Example Question #1 : Rotational Motion And Torque

Example Question #54 : Ap Physics C

Example Question #55 : Ap Physics C

All AP Physics C: Mechanics Resources

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