Question

I keep getting the wrong answer help :/
A barbell spins around a pivot at its center
at A. The barbell consists of two small balls,
each with mass m = 0.3 kg, at the ends of a
massless rod of length d = 0.7 m. The barbell
spins clockwise on the plane of the paper with
an angular speed !0 = 20 rad/s (choice of
axes: x to the right, y up, z out of the page;
origin at A).

Calculate the z component of the total angular momentum for this system of two particles.
Answer in units of kg · m2/s

Answer 1

and I seem to have pasted it on top of itself. Also sorry.

Answer 2

\[\omega = 20 rad/s=\frac{v}{r}\]

** (see the bottom for less complication)
Since everything is constant, we should be able to solve at a specific time to make it easier

velocity of "top" barbel at y axis is in the negative x direction
\[\textbf v_{top} = -(.35m)(20rad/s) \hat{ \textbf x}=-7m/s \ \hat{ \textbf x}\]

and the velocity of the "bottom" barbel at the y axis is in the positive x direction
\[\textbf v_{bot} = 7m/s \ \hat{ \textbf x}\]

Noting that
\[\textbf r_{top} = .35m \ \hat{ \textbf y}
\\ \
\\ \textbf r_{top} = -.35m \ \hat{ \textbf y}\]

Then
\[ \sum \textbf L = \sum_i \textbf r_i \times m_i \textbf v_i
\\ \
\\
\hspace{33px}= \textbf r_{top} \times m_1 \textbf v_{top} + \textbf r_{bot} \times m_2 \textbf v_{bot}
\\
\\
\hspace{33px} = m_1\big(\textbf r_{top} \times \textbf v_{top} \big) + m_2\big( \textbf r_{bot} \times \textbf v_{bot} \big)
\]

So you set up two cross products

\[\textbf r_{top} \times \textbf v_{top} =
\[\omega = 20 rad/s=\frac{v}{r}\]

Since everything is constant, we should be able to solve at a specific time to make it easier

velocity of "top" barbel at y axis is in the negative x direction
\[\textbf v_{top} = -(.35m)(20rad/s) \hat{ \textbf x}=-7m/s \ \hat{ \textbf x}\]

and the velocity of the "bottom" barbel at the y axis is in the positive x direction
\[\textbf v_{bot} = 7m/s \ \hat{ \textbf x}\]

Noting that
\[\textbf r_{top} = .35m \ \hat{ \textbf y}
\\ \
\\ \textbf r_{top} = -.35m \ \hat{ \textbf y}\]

Then
\[ \sum \textbf L = \sum_i \textbf r_i \times m_i \textbf v_i
\\ \
\\
\hspace{33px}= \textbf r_{top} \times m_1 \textbf v_{top} + \textbf r_{bot} \times m_2 \textbf v_{bot}
\\
\\
\hspace{33px} = m_1\big(\textbf r_{top} \times \textbf v_{top} \big) + m_2\big( \textbf r_{bot} \times \textbf v_{bot} \big)
\]

So you set up two cross products

\[\textbf r_{top} \times \textbf v_{top} = \begin{vmatrix} x & y & z \\ r_x & r_y & r_z \\ v_x & v_y & v_z \end{vmatrix} = \begin{vmatrix} x & y & z \\ 0 & .35 & 0 \\ -7 & 0 & 0 \end{vmatrix}=2.45m^2/s \ \hat{ \textbf z}\]

The bottom will give you the same thing, so the total angular momentum is

\[ \textbf L = (m_1+m_2)2.45m^2/s \ \ \hat{ \textbf z}\]

Which I totally just now realize is is exactly what you would get if you did it *not this super complicated way and just added up the two using the basic formula for moment of inertia of a point mass about an axis and angular momentum

\[I = mr^2\]

\[L = I \omega\]

\[ \sum L = I_1\omega + I_2\omega = m_1r^2\omega + m_2r^2\omega = (m_1+m_2)(r^2\omega)\]

sorry bout that.