Question

AP Physics C: Center of Mass and Linear Momentum. Please help!

Answer 1

The figure shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass m1 = 0.700 kg, and its center is initially at xy coordinates (–0.400 m, 0 m); the block has mass m2 = 0.350 kg, and its center is initially at xy coordinates (0, –0.210 m). The mass of the cord and pulley are negligible. The cart is released from rest, and both cart and block move until the cart hits the pulley. The friction between the cart and the air track and between the pulley and its axle is negligible. (a) In unit-vector notation, what is the acceleration of the center of mass of the cart–block system? (b) What is the velocity of the com as a function of time t, in unit-vector notation?

picture: http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/ch0/EAT_13439217832640_6624324733191566.gif

Answer 2

@amistre64

Answer 3

So do you know the vector equation to find the center of mass of a system?

Answer 4

(the notification system has been acting up lately, so if the folks you're tagging aren't responding, it's because it's not going through ^_^)

Answer 5

In any case, to find the center of mass

\[ \textbf R_{CoM}=\frac{1}{M}\sum_{\alpha}m_{\alpha}\textbf r_{\alpha}\]
where M is the sum of all the masses in the system.

Answer 6

In this case, we would have
\[ \textbf R = \frac{m_1 \textbf r_1 + m_2 \textbf r_2}{m_1+m_2}\]

\[ \text{Can you express } \textbf r_1 \text{ and } \textbf r_2 in unit vector notation? (with i, j, and k hats)?

Answer 7

\[ \text{Can you express } \textbf r_1 \text{ and } \textbf r_2 \text{ in unit vector notation? (with i, j, and k hats)?}\]

Answer 8

\[r_1=-0.4i\]\[r_2=-0.21j\]I think?

Answer 9

is that right?

Answer 10

correct! Now, looking at that equation for the center of mass, if we take the first and second derivatives (without plugging in values), we can see something cool

\[ \frac{ \mathrm{d}\textbf R_{CoM}}{\mathrm{d}t} = \frac{m_1 \frac{ \mathrm{d} \textbf r_1}{\mathrm{d}t}+ m_2\frac{ \mathrm{d} \textbf r_2}{\mathrm{d}t}}{m_1+m_2}\]
\[ \textbf v_{CoM} = \frac{m_1 \textbf v_1 + m_2 \textbf v_2}{m_1+m_2}\]

and similarly
\[ \frac{ \mathrm{d}^2\textbf R_{CoM}}{\mathrm{d}t^2} = \frac{m_1 \frac{ \mathrm{d}^2 \textbf r_1}{\mathrm{d}t^2}+ m_2\frac{ \mathrm{d}^2 \textbf r_2}{\mathrm{d}t^2}}{m_1+m_2}\]
\[ \textbf a_{CoM} = \frac{m_1 \textbf a_1 + m_2 \textbf a_2}{m_1+m_2}\]

So to find the acceleration of the center of mass, we can just each of the vectors and add em up! So can you find the acceleration on each block separately, like a normal free body diagram problem?

Answer 11

Well, I came up with these equations:\[m_2g-T=m_2a\]\[Normal=m_1g\]\[T-f_k=m_1a\]but I don't think this is enough info to solve for acceleration

Answer 12

luckily we don't have to deal with friction in this problem, and there are enough equations there to solve for acceleration! We can equate the T's, because tension will be the same on both sides of a frictionless massless pulley.


\[T=m_1a\]
\[ m_2g-T = m_2a\]
\[m_2g - m_1a = m_2a\]
\[a=g\frac{m_2}{m_1+m_2}\]

That's the magnitude of the acceleration - can you put that into 2 more unit vector notation forms for each mass?

Answer 13

Just remember that
\[\textbf a_1 \text{ is purely in the positive x direction}\]
and that

\[ \textbf a_2 \text{ is purely in the negative y direction}\]


Similarly for velocity, you can use Vi = 0, the acceleration of each block, and kinematics to come up with an expression in terms of t. Again, the velocity of mass 1 will be purely in the positive x direction, and the velocity of mass 2 will be purely in the negative y direction.

Also good to remember, for adding vectors, if

\[\textbf r_1 = a \ \hat{\textbf i} + b \ \hat{\textbf j}\]
\[\textbf r_2 = c \ \hat{\textbf i} - d \ \hat{\textbf j}\]
\[ \textbf r = \textbf r_1 + \textbf r_2
\\ \
\\ = (a+c) \ \hat{\textbf i} + (b-d) \ \hat{\textbf j}\]