Question

The uniform bar shown in the diagram has a length of 0.80 m. The bar begins to rotate from rest in the horizontal plane about the axis passing through its left end. What will be the magnitude of the angular momentum L of the bar 6.0 s after the motion has begun? The forces acting on the bar are shown.

Express your answer in kg*m2/s

Answer 1

Concepts and reason

The concepts required to solve the given questions are torque and the angular momentum.

Initially, calculate the net torque. Finally, calculate the final angular momentum.

Fundamentals

The expression for the torque is as follows:

$\tau = \frac{{d\vec L}}{{dt}}$

Here, $d\vec L$ is the change in the angular momentum and dt is the change in the time.

Also, the expression for the torque is as follows:

$\tau = \vec r \times \vec F$

Here, $\vec r$ is the position vector and $\vec F$ is the force.

The expression for the angular momentum is as follows:

$L = \vec r \times \vec p$

Here, $\vec p$ is the linear momentum vector and $\vec r$ is the position vector.

The net torque acting is as follows:

${\tau _{{\rm{net}}}} = {\vec r_1} \times {\vec F_1} + {\vec r_2} \times {\vec F_2}$

Here, ${\vec r_1}$ is the distance from the axis to the first force ${\vec F_1}$ and ${\vec r_2}$ is the distance from the axis to the second force ${\vec F_2}$ .

Substitute $0.60{\rm{ m}}\left( {\hat x} \right)$ for ${\vec r_1}$ , $12{\rm{ N}}\left( {\hat y} \right)$ for ${\vec F_1}$ , $0.80{\rm{ m}}\left( {\hat x} \right)$ for ${\vec r_2}$ , and ${\rm{8 N}}\left( { - \hat y} \right)$ for ${\vec F_2}$ in the equation ${\tau _{{\rm{net}}}} = {\vec r_1} \times {\vec F_1} + {\vec r_2} \times {\vec F_2}$ .

$\begin{array}{c}\\{\tau _{{\rm{net}}}} = \left( {0.60{\rm{ m}}\left( {\hat x} \right)} \right) \times \left( {12{\rm{ N}}\left( {\hat y} \right)} \right) + \left( {0.80{\rm{ m}}\left( {\hat x} \right)} \right) \times \left( {{\rm{8 N}}\left( { - \hat y} \right)} \right)\\\\ = 0.80{\rm{ N}} \cdot {\rm{m}}\left( {\hat z} \right)\\\end{array}$

Therefore, the magnitude of the net torque is as follows:

$\left| {{\tau _{{\rm{net}}}}} \right| = 0.8{\rm{0 N}} \cdot {\rm{m}}$

From the definition of torque,

$\begin{array}{c}\\\tau = \frac{{d\vec L}}{{dt}}\\\\ = \frac{{{{\vec L}_{\rm{f}}} - {{\vec L}_{\rm{i}}}}}{{\left( {{t_{\rm{f}}} - {t_{\rm{i}}}} \right)}}\\\end{array}$

Therefore, the expression for torque is as follows:

$\tau = \frac{{{{\vec L}_{\rm{f}}} - {{\vec L}_{\rm{i}}}}}{{\left( {{t_{\rm{f}}} - {t_{\rm{i}}}} \right)}}$

Here, suffix i represents the initial and f represents the final.

The expression for the initial angular momentum is given by,

${\vec L_{\rm{i}}} = {\vec r_{\rm{i}}} \times {\vec p_{\rm{i}}}$

Here, ${\vec r_{\rm{i}}}$ is the initial position vector and ${\vec p_{\rm{i}}}$ is the initial momentum vector.

Now, substitute ${\vec r_{\rm{i}}} \times {\vec p_{\rm{i}}}$ for ${\vec L_{\rm{i}}}$ in the equation $\tau = \frac{{{{\vec L}_{\rm{f}}} - {{\vec L}_{\rm{i}}}}}{{\left( {{t_{\rm{f}}} - {t_{\rm{i}}}} \right)}}$ .

$\begin{array}{c}\\\tau = \frac{{{{\vec L}_{\rm{f}}} - \left( {{{\vec r}_{\rm{i}}} \times {{\vec p}_{\rm{i}}}} \right)}}{{\left( {{t_{\rm{f}}} - {t_{\rm{i}}}} \right)}}\\\\ = \frac{{{{\vec L}_{\rm{f}}} - \left( {{{\vec r}_{\rm{i}}} \times m{{\vec v}_{\rm{i}}}} \right)}}{{\left( {{t_{\rm{f}}} - {t_{\rm{i}}}} \right)}}\\\end{array}$

Substitute 0.00 m/s for ${\vec v_{\rm{i}}}$ and $0.00{\rm{ s}}$ for ${t_{\rm{i}}}$ in above equation and rearrange the equation as follows:

$\begin{array}{c}\\\tau = \frac{{{{\vec L}_{\rm{f}}} - m\left( {{{\vec r}_{\rm{i}}} \times \left( {0.00{\rm{ m/s}}} \right)} \right)}}{{\left( {{t_{\rm{f}}} - \left( {0.00{\rm{ s}}} \right)} \right)}}\\\\0.80{\rm{ N}} \cdot {\rm{m}} = \frac{{{{\vec L}_{\rm{f}}}}}{{{t_{\rm{f}}}}}\\\\{{\vec L}_{\rm{f}}} = {t_{\rm{f}}}\left( {0.80{\rm{ N}} \cdot {\rm{m}}} \right)\\\end{array}$

Substitute 6.0 s for ${t_{\rm{f}}}$ in the equation ${\vec L_{\rm{f}}} = {t_{\rm{f}}}\left( {0.80{\rm{ N}} \cdot {\rm{m}}} \right)$ .

$\begin{array}{c}\\{{\vec L}_{\rm{f}}} = \left( {6.0{\rm{ s}}} \right)\left( {0.80{\rm{ N}} \cdot {\rm{m}}} \right)\\\\ = 4.8{\rm{ kg}} \cdot {{\rm{m}}^2} \cdot {{\rm{s}}^{ - 1}}\\\end{array}$

Ans:

The value of the magnitude of the angular momentum is equal to $4.8{\rm{ kg}} \cdot {{\rm{m}}^2} \cdot {{\rm{s}}^{ - 1}}$ .