In figure a barbell spins around a pivot at its center at\(A\). The barbell consists of two small balls, each with mass \(500{\rm{ g}}\left( {0.5{\rm{ kg}}} \right)\), at the ends of a very low mass rod of length the \(d = 20{\rm{ cm}}\left( {0.2{\rm{ m}}} \right)\) the radius of rotation is\(0.1{\rm{ m}}\)). The barbell spins clockwise with angular speed \(80{\rm{ }}{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.s} {\rm{}}}\)

We can calculate the angular momentum and kinetic energy of this object in two different ways, by treating the object as two separate balls or as one barbell. Use the usual coordinate system, with\(x\)to the right,\(y\)toward the top of the page, and\(z\)out of the page, toward you.

I . Treat the object as two separate balls. Calculate the following quantities:

a.The speed of ball 1, (b)\({\overrightarrow L _{trans,1,A}}\)of ball 1, (c)\({\overrightarrow L _{trans,2,A}}\)of ball 2, (d)\({\overrightarrow L _{tot,A}}\)(e) the translational kinetic energy of ball 1, (f) the translational kinetic energy of ball 3, (g) the total kinetic energy of the barbell.

II .Treat the object as one barbell. Calculate the following quantities:

(h) The moment of inertia\(I\)of the barbell, (i)\(\overrightarrow \omega \)expressed as a vector, (j)\({\overrightarrow L _{rot}}\)of the barbell, (k)\({K_{rot}}\).

III. Compare the two approaches:

1. Compare your result for\({\overrightarrow L _{tot,A}}\)in part\(I\)to your result for\({\overrightarrow L _{rot}}\)in part\(II\). Should these quantities be the same or different?
2. Compare your result for\({K_{total}}\)in part\(I\)to your result for\({K_{rot}}\)in part\(II\). Should these quantities be the same, or different?

Angular momentum, property characterizing the rotary inertia of an object or system of objects in motion about an axis that may or may not pass through the object or system.

Moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

I.

Use the radius of rotation and energy velocity of the ball 1 to find the speed of the ball 1

The figure shows that a barbell spins around a pivot point at point A.

The speed of the ball 1 in terms of angular speed is,

\({v_1} = {r_1}{\omega _1}\)

Here, \({r_1}\)is radius of rotation of ball 1 and \({\omega _1}\)is angular speed of the ball 1.

Substitute \(10{\rm{ cm}}\) for \({r_1}\) and \(18{\rm{ }}{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\) for \({\omega _1}\).

\(\begin{aligned}{}{v_1} &= \left( {10{\rm{ cm}}} \right)\left( {18{\rm{ }}{{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\ &= \left( {10{\rm{ cm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\left( {18{\rm{ }}{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\ &= 8{\rm{ }}{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Hence, the speed of ball 1 is\(8{\rm{ }}{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}\).

The linear momentum of the ball 1 is,

\({p_1} = {m_1}{v_1}\)

Here, \({m_1}\) is mass of the ball 1.

Substitute \(0.5{\rm{ kg}}\) for \({m_1}\) and \(8{\rm{ }}{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\)for \({v_1}\) in the above equation.

\(\begin{aligned}{}{p_1} = \left( {0.5{\rm{ kg}}} \right)\left( {8{\rm{ }}{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\ = 4.0{\rm{ }}{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/ {\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

The transitional angular momentum of the ball 1 with respect to A is

\({\overrightarrow L _{trans,1,A}} = {r_1}{p_1}\sin {\theta _1}\)

Here, \({\theta _1}\)is angle between \({r_1}\)and \({p_1}\).

Substitute \(10{\rm{ cm}}\)for \({r_1}\), \(4.0{\rm{ }}{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/

{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.} {\rm{s}}}\) for \({p_1}\), and \(90^\circ \)for \({\theta _1}\) in the above equation.

\(\begin{aligned}{}{\overrightarrow L _{trans,1,A}} = \left( {10{\rm{ cm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ m}}}}} \right)\left( {4.0{\rm{ }}{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/

{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\sin 90^\circ \\ = 0.40{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} \mathord{\left/

{\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Therefore, the transitional angular momentum of the ball 1 with respect to A is \(0.40{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {\rm{s}}}} \right.} {\rm{s}}}\)and it is directed along normally inward to the plane of paper.

The speed of the ball 2 in terms of angular speed is,

\({v_2} = {r_2}{\omega _2}\)

Here, \({r_2}\)is radius of rotation of ball 2 and\({\omega _2}\)is angular speed of the ball 2.

Substitute \(10{\rm{ cm}}\)for \({r_2}\)and \(80{\rm{ }}{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.}{\rm{s}}}\)for \({\omega _2}\).

\(\begin{aligned}{}{v_2} = \left( {10{\rm{ cm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\left( {80{\rm{ }}{{{\rm{rad}}} \mathord{\left/{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\ = 8.0{\rm{ }}{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

The linear momentum of the ball 2 is,

\({p_2} = {m_2}{v_2}\)

Here, \({m_2}\) is mass of the ball 2.

Substitute \(0.5{\rm{ kg}}\)for \({m_2}\)and \(8.0{\rm{ }}{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\)for \({v_2}\) in the above equation, and you get

\(\begin{aligned}{c}{p_2} = \left( {0.5{\rm{ kg}}} \right)\left( {8.0{\rm{ }}{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\ = 4.0{\rm{ }}{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

The transitional angular momentum of the ball 2 with respect to A is,

\({\overrightarrow L _{{\mathop{\rm t}\nolimits} rans,2,A}} = {r_2}{p_2}\sin {\theta _2}\)

Here, \({\theta _2}\)is angle between \({r_2}\)and \({p_2}\).

Put\(10{\rm{ cm}}\) for \({r_2}\), \(4.0{\rm{ }}{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/

{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.} {\rm{s}}}\) for\({p_2}\), and \(90^\circ \)for \({\theta _2}\) in the above equation.

\(\begin{aligned}{c}{\overrightarrow L _{{\mathop{\rm t}\nolimits} rans,2,A}} = \left( {10{\rm{ cm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)\left( {4.0{\rm{ }}{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/ {\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\sin 90^\circ \\ = 0.40{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} \mathord{\left/

{\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Hence, the transitional angular momentum of the ball 2 with respect to A to \(0.40{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} \mathord{\left/

{\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {\rm{s}}}} \right.} {\rm{s}}}\)and it is directed along normally inward to the plane of paper.

The total transitional angular momentum of the system of the two balls is,

\({\overrightarrow L _{tot,A}} = {\overrightarrow L _{trans,1,A}} + {\overrightarrow L _{trans,2,A}}\)

Substitute \(0.40{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} \mathord{\left/

{\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {\rm{s}}}} \right.} {\rm{s}}}\)for \({\overrightarrow L _{trans,1,A}}\) and \(0.40{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} \mathord{\left/

{\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {\rm{s}}}} \right.} {\rm{s}}}\)for \({\overrightarrow L _{trans,2,A}}\)

\(\begin{aligned}{c}{\overrightarrow L _{tot,A}} = 0.40{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} \mathord{\left/ {\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {\rm{s}}}} \right.} {\rm{s}}} + 0.40{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} \mathord{\left/

{\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {\rm{s}}}} \right.} {\rm{s}}}\\ = 0.80{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} \mathord{\left/

{\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Therefore, the total transitional angular momentum of the system of the two balls is and \(0.80{\rm{ }}{{{\rm{kg}} \cdot {{\rm{m}}^2}} \mathord{\left/

{\vphantom {{{\rm{kg}} \cdot {{\rm{m}}^2}} {\rm{s}}}} \right.} {\rm{s}}}\)is directed normally into the page.