Question

Learning Goal: To understand and apply the formula τ=Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle of mass m, we use Newton's second law: F⃗ net=ma⃗ , where F⃗ net is the net force acting on the particle. To find the angular acceleration α of a rigid object rotating about a fixed axis, we can use a similar formula: τnet=Iα, where τnet=∑τ is the net torque acting on the object and I is its moment of inertia.

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction.Assume that the pivot is attached tot he center of the bar. You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2. Part A Assume that the mass of the swing bar, as shown in the figure, is negligible.(Figure 1) Find the magnitude of the angular acceleration α of the seesaw. Express your answer in terms of some or all of the quantities m1, m2, l, as well as the acceleration due to gravity g. α = 2(m1g−m2g)1(m1+m2) SubmitHintsMy AnswersGive UpReview Part Incorrect; Try Again; 6 attempts remaining Part B In what direction will the seesaw rotate, and what will the sign of the angular acceleration be? The rotation is in the clockwise direction and the angular acceleration is positive. The rotation is in the clockwise direction and the angular acceleration is negative. The rotation is in the counterclockwise direction and the angular acceleration is positive. The rotation is in the counterclockwise direction and the angular acceleration is negative. SubmitMy AnswersGive Up Correct Part C Now consider a similar situation, except that now the swing bar itself has mass mbar.(Figure 2) Find the magnitude of the angular acceleration α of the seesaw. Express your answer in terms of some or all of the quantities m1, m2, mbar, l, as well as the acceleration due to gravity g. α = SubmitHintsMy AnswersGive UpReview Part Part D In what direction will the seesaw rotate and what will the sign of the angular acceleration be? In what direction will the seesaw rotate and what will the sign of the angular acceleration be? The rotation is in the clockwise direction and the angular acceleration is positive. The rotation is in the clockwise direction and the angular acceleration is negative. The rotation is in the counterclockwise direction and the angular acceleration is positive. The rotation is in the counterclockwise direction and the angular acceleration is negative.

Answer 1

Concepts and reason

The concept required to solve this problem is torque and moment of inertia.

First, calculate the magnitude of the angular acceleration of the seesaw if the mass of the swing bar is negligible by using the moment of inertia and torque. Next, find out the direction and sign of the angular acceleration by using the expression for the angular acceleration. After that, calculate the magnitude of the angular acceleration of the seesaw by using the moment of inertia and torque. Finally, find out the direction and sign of the angular acceleration by using the expression for the angular acceleration.

Fundamentals

The torque is given as,

$\tau = I\alpha$

Here, I is the total moment inertia of the object and $\alpha$ is the angular acceleration.

The net torque is given by the sum of all the torques.

${\tau _{{\rm{net}}}} = \sum {rF}$

Here, F is the force on the object and r is the perpendicular distance from the axis of rotation.

The moment of inertia of on object is,

$I = m{r^2}$

Here, $m$ is the mass, and $r$ is the perpendicular distance from the axis of rotation.

The moment of inertia of the rod about the passing through its center is,

${I_{{\rm{rod}}}} = \frac{1}{{12}}{m_{{\rm{rod}}}}{L^2}$

Here, ${m_{{\rm{rod}}}}$ is the mass of the rod and $L$ is the length of the rod.

The force of gravity or weight of an object is,

${F_{\rm{g}}} = mg$

Here, $m$ is the mass, and $g$ is the acceleration due to gravity.

Sign convention used is as follows:

All the forces downward are negative and upward are positive.

All the anticlockwise torques are taken as positive and clockwise torques are taken as negative.

Part A

Substitute $\frac{L}{2}$ for $r$, ${m_1}$ for $m$, and ${I_1}$ for $I$ in the equation $I = m{r^2}$ to solve for the moment of inertia of the mass ${m_1}$.

$\begin{array}{c}\\{I_1} = {m_1}{\left( {\frac{L}{2}} \right)^2}\\\\ = \frac{{{m_1}{L^2}}}{4}\\\end{array}$

Substitute $\frac{L}{2}$ for $r$, ${m_2}$ for $m$, and ${I_2}$ for $I$ in the equation $I = m{r^2}$ to solve for the moment of inertia of the mass ${m_2}$.

$\begin{array}{c}\\{I_2} = {m_2}{\left( {\frac{L}{2}} \right)^2}\\\\ = \frac{{{m_2}{L^2}}}{4}\\\end{array}$

The total moment of inertia of the bar and objects is sum of the moment of inertia’s that is,

$I = {I_1} + {I_2}$

Substitute $\frac{{{m_1}{L^2}}}{4}$ for ${I_1}$, and $\frac{{{m_2}{L^2}}}{4}$ for ${I_2}$ in the equation $I = {I_1} + {I_2}$.

$\begin{array}{c}\\I = \frac{{{m_1}{L^2}}}{4} + \frac{{{m_2}{L^2}}}{4}\\\\ = \left( {{m_1} + {m_2}} \right)\left( {\frac{{{L^2}}}{4}} \right)\\\end{array}$

Use the net torque equation.

Substitute ${r_1}{F_1} - {r_2}{F_2}$ for $\sum {rF}$, and $I\alpha$ for ${\tau _{{\rm{net}}}}$ in the net torque equation about the center of the swing bar ${\tau _{{\rm{net}}}} = \sum {rF}$.

$I\alpha = {r_1}{F_1} - {r_2}{F_2}$

Substitute $\left( {{m_1} + {m_2}} \right)\left( {\frac{{{L^2}}}{4}} \right)$ for $I$, $\frac{L}{2}$ for ${r_1}$, ${m_1}g$ for ${F_1}$, $\frac{L}{2}$ for ${r_2}$, and ${m_2}g$ for ${F_2}$ in the equation $I\alpha = {r_1}{F_1} - {r_2}{F_2}$ and solve for the angular acceleration $\alpha$.

$\begin{array}{c}\\\left( {{m_1} + {m_2}} \right)\left( {\frac{{{L^2}}}{4}} \right)\alpha = \left( {\frac{L}{2}} \right)\left( {{m_1}g} \right) - \left( {\frac{L}{2}} \right)\left( {{m_2}g} \right)\\\\\alpha = \frac{{{m_1}g - {m_2}g}}{{\left( {{m_1} + {m_2}} \right)\left( {\frac{L}{2}} \right)}}\\\\ = 2\left( {\frac{g}{L}} \right)\left( {\frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}}} \right)\\\end{array}$

Part B

The angular acceleration of the swing bar is given as,

$\alpha = 2\left( {\frac{g}{L}} \right)\left( {\frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}}} \right)$

The given condition between the masses is, ${m_1} > {m_2}$.

If ${m_1} > {m_2}$ then ${m_1} - {m_2} = {\rm{positive}}$. The angular acceleration is positive that is it is in counterclockwise direction. Thus, the rotation is in counterclockwise direction.

Part C

Use the moment of inertia equation.

Substitute $\frac{L}{2}$ for $r$, ${m_1}$ for $m$, and ${I_1}$ for $I$ in the equation $I = m{r^2}$ to solve for the moment of inertia of the mass ${m_1}$.

$\begin{array}{c}\\{I_1} = {m_1}{\left( {\frac{L}{2}} \right)^2}\\\\ = \frac{{{m_1}{L^2}}}{4}\\\end{array}$

Substitute $\frac{L}{2}$ for $r$, ${m_2}$ for $m$, and ${I_2}$ for $I$ in the equation $I = m{r^2}$ to solve for the moment of inertia of the mass ${m_2}$.

$\begin{array}{c}\\{I_2} = {m_2}{\left( {\frac{L}{2}} \right)^2}\\\\ = \frac{{{m_2}{L^2}}}{4}\\\end{array}$

The total moment of inertia of the bar and objects is sum of the moment of inertia’s that is,

$I = {I_1} + {I_2} + {I_{{\rm{rod}}}}$

Substitute $\frac{{{m_1}{L^2}}}{4}$ for ${I_1}$, $\frac{{{m_2}{L^2}}}{4}$ for ${I_2}$, and $\frac{1}{{12}}{m_{{\rm{rod}}}}{L^2}$ for in the equation $I = {I_1} + {I_2}$.

$\begin{array}{c}\\I = \frac{{{m_1}{L^2}}}{4} + \frac{{{m_2}{L^2}}}{4} + \frac{1}{{12}}{m_{{\rm{rod}}}}{L^2}\\\\ = \left( {{m_1} + {m_2} + \frac{{{m_{{\rm{rod}}}}}}{3}} \right)\left( {\frac{{{L^2}}}{4}} \right)\\\end{array}$

Use the net torque equation.

Substitute ${r_1}{F_1} - {r_2}{F_2}$ for $\sum {rF}$, and $I\alpha$ for ${\tau _{{\rm{net}}}}$ in the net torque equation about the center of the swing bar ${\tau _{{\rm{net}}}} = \sum {rF}$.

$I\alpha = {r_1}{F_1} - {r_2}{F_2}$

Substitute $\left( {{m_1} + {m_2} + \frac{{{m_{{\rm{rod}}}}}}{3}} \right)\left( {\frac{{{L^2}}}{4}} \right)$ for $I$, $\frac{L}{2}$ for ${r_1}$, ${m_1}g$ for ${F_1}$, $\frac{L}{2}$ for ${r_2}$, and ${m_2}g$ for ${F_2}$ in the equation $I\alpha = {r_1}{F_1} - {r_2}{F_2}$ and solve for the angular acceleration $\alpha$.

$\begin{array}{c}\\\left( {{m_1} + {m_2} + \frac{{{m_{{\rm{rod}}}}}}{3}} \right)\left( {\frac{{{L^2}}}{4}} \right)\alpha = \left( {\frac{L}{2}} \right)\left( {{m_1}g} \right) - \left( {\frac{L}{2}} \right)\left( {{m_2}g} \right)\\\\\alpha = \frac{{{m_1}g - {m_2}g}}{{\left( {{m_1} + {m_2} + \frac{{{m_{{\rm{rod}}}}}}{3}} \right)\left( {\frac{L}{2}} \right)}}\\\\ = 2\left( {\frac{g}{L}} \right)\left( {\frac{{{m_1} - {m_2}}}{{{m_1} + {m_2} + \frac{{{m_{{\rm{rod}}}}}}{3}}}} \right)\\\end{array}$

Part D

The angular acceleration of the swing bar is given as,

$\alpha = 2\left( {\frac{g}{L}} \right)\left( {\frac{{{m_1} - {m_2}}}{{{m_1} + {m_2} + \frac{{{m_{{\rm{rod}}}}}}{3}}}} \right)$

The given condition between the masses is, ${m_1} > {m_2}$.

If ${m_1} > {m_2}$ then ${m_1} - {m_2} = {\rm{positive}}$. The angular acceleration is positive that is it is in counterclockwise direction. Thus, the rotation is in counterclockwise direction.

Ans: Part A

The magnitude of angular acceleration of the swing bar is $\alpha = 2\left( {\frac{g}{L}} \right)\left( {\frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}}} \right)$.