91Ó°ÊÓ

The kinetic energy is the measure of the work that an object does by virtue of its motion. Simple act like walking, jumping, throwing, and falling involves kinetic energy.

The momentum of translation being a vector quantity in classical physics equal to the product of the mass and the velocity of the centre of mass.

Apply conservation of angular momentum to solve for the final angular velocity of the play-ground ride by considering the child and the play-ground ride as a system.

Initial angular momentum of the child is,$L_{i,child}=mvR$

Initially the disk is at rest, so the initial angular momentum of the disk is,

$L_{i,disk}=0$

The initial angular momentum of the system is,

$L_i=mvR$

The moment of inertia of the playground (disk) relative to the axle is,

$I_{play}=\frac{1}{2}M{R}^2$

The moment of inertia of the child relative to the axle is the sum of the moment of inertias of the child and the playground.

$I_f=I_{child}+I_{play}$

$$\begin{gathered} =m{R}^2+\frac{1}{2}M{R}^2 \\ =\left(m+\frac{M}{2}\right)R^2 \end{gathered}$$

The final angular momentum of the system is

$L_f=I_f+{\mathrm{Ó\neg }}_f$

The net external torque acting on the system is zero, so the angular momentum of the system is conserved.

$$\begin{gathered} {\stackrel{→}{\mathrm{τ}}}_{net}=\frac{d\stackrel{→}{L}}{dt} \\ 0=\frac{d\stackrel{→}{L}}{dt} \\ \stackrel{→}{L_f}={\stackrel{→}{L}}_i \end{gathered}$$

Therefore, the final angular speed of the disk can be calculated as,

$$\begin{gathered} \mathrm{Ó\neg }=\frac{L_f}{I_f} \\ =\frac{mvR}{\left(m+\frac{M}{2}\right)R^2} \\ =\frac{mv}{\left(m+\frac{M}{2}\right)R} \end{gathered}$$

Substitute $25kg$for$m,2.3m/s$for$v,43kg$for $M$and $1.7m$for $R$in $\mathrm{Ó\neg }=\frac{mv}{\left(m+\frac{M}{2}\right)R}$

$\mathrm{Ó\neg }=\frac{\left(25kg\right)(2.3m/s)}{\left(25kg+\frac{43kg}{2}\right)(1.7m)}$

$=0.727rad/s$

Thus, the final angular velocity of the disk is$0.727rad/s$

Substitute $25kg$for $m,$and $2.3m/s$f

$$\begin{gathered} =K_C=\frac{1}{2}\left(25kg\right)(2.3m/s{)}^2 \\ =66.1J \end{gathered}$$

Initially the disk is at rest, so the initial angular speed of the disk is$0.727rad/s$

The initial rotational kinetic energy of the system is zero.

Therefore, the total initial kinetic energy of the child-disk system is,

$K_i=66.1J$

Final angular speed of the disk,${\mathrm{Ó\neg }}_f=0.727rad/s$

Final, moment of inertia of the system,$I_f\left(m+\frac{M}{2}\right)R^2$

The final kinetic energy of the system can be calculated as

$$\begin{gathered} K_f=\frac{1}{2}{I}_f{{\mathrm{Ó\neg }}_f}^2 \\ =\frac{1}{2}\left(m+\frac{M}{2}\right)R^2{{\mathrm{Ó\neg }}_f}^2 \\ =\frac{1}{2}\left(25kg+\frac{43kg}{2}\right)(1.7m{)}^2(0.727rad/s{)}^2 \\ =35.51J \end{gathered}$$

Therefore, the change in the kinetic energy of the child plus disk is,

$$\begin{gathered} \mathrm{Δ}K=K_f-K_i \\ =35.51J-66.1J \\ =-30.6J \end{gathered}$$

Thecomponent of the initial linear momentum of the system is

$$\begin{gathered} =p_{i,x}=mv \\ =\left(25kg\right)(23m/s \\ =57.5kg.m/s \end{gathered}$$

The component of the final linear momentum of the system is

$p_{f,x}=84.048kg.m/s$

Therefore, the change in thecomponent of linear momentumis

$$\begin{gathered} \mathrm{Δ}{p}_x=p_{f,x}=p_{i,x} \\ =84.048kg.m/s-57.5kg.m/s \\ =26.548kg.m/s \end{gathered}$$

The linear momentum of the system changed due to the force exerted by the axle on it.

Conserve angular momentum as there is no external torques acting on the system. The expression for the initial moment of inertia of the system is,

$$\begin{gathered} I_i=m{R}^2+\frac{1}{2}M{R}^2 \\ =\left(m+\frac{M}{2}\right)R^2 \end{gathered}$$

Substitute$25kg$for$m,43kg$for$M,$and$1.7m$for $R$in$I_i=\left(m+\frac{M}{2}\right)R^2$

$$\begin{gathered} I_i=\left(25kg+\frac{43kg}{2}\right)(1.7m{)}^2 \\ =134.385kg.m^2 \end{gathered}$$

The final moment of inertia of the child at a distancerelative to the axle is the sum of the moment of inertias of the child and the playground.

$$\begin{gathered} I_f=I_{child}+I_{play} \\ =m(R\text{'}{)}^2+\frac{1}{2}M{R}^2 \\ =\left(25kg\right)(0.7m{)}^2+\frac{1}{2}\left(43kg\right)(1.7m{)}^2 \\ =74.385kg.m^2 \end{gathered}$$

Apply the law of conservation of angular momentum to the system.

$I_i{\mathrm{Ó\neg }}_i=I_f+{\mathrm{Ó\neg }}_f$

The final angular speed of the system can be calculated as

$$\begin{gathered} {\mathrm{Ó\neg }}_f=\left(\frac{I_i}{I_f}\right){\mathrm{Ó\neg }}_i \\ =\left(\frac{134.385kg.m^2}{74.385kg.m^2}\right)(0.727rad/s) \\ =1.313rad \end{gathered}$$

$$\begin{gathered} {\mathrm{Ó\neg }}_f=\left(\frac{I_i}{I_f}\right){\mathrm{Ó\neg }}_i \\ =\left(\frac{134.385kg.m^2}{74.385kg.m^2}\right)(0.727rad/s) \\ =1.313rad/s \end{gathered}$$

$k_f=\frac{1}{2}{I}_f{{\mathrm{Ó\neg }}_f}^2$

Substitute$74.385kg.m$for$I_f$and$1.313rad/s$for ${\mathrm{Ó\neg }}_f$in $k_f=\frac{1}{2}{I}_f{{\mathrm{Ó\neg }}_f}^2.$

$k_f=\frac{1}{2}{I}_f{{\mathrm{Ó\neg }}_f}^2.$

$$\begin{gathered} k_f=\frac{1}{2}(74.385kg.m^2)(1.313rad/s{)}^2 \\ =64.11J \end{gathered}$$

Thus, change in kinetic energy is,

$$\begin{gathered} k_f-k_i=64.11J-35.51J \\ =29J \end{gathered}$$

Therefore, the change in kinetic energy is$28.6J$