Question

A playground ride consists of a disk of mass $M=43kg$and radius $R=1.7m$mounted on a low-friction axle (Figure 11.94). A child of mass $m=25kg$runs at speed $v=2.3m/s$on a line tangential to the disk and jumps onto the outer edge of the disk.

(a.) If the disk was initially at rest, now how fast is it rotating? (b) What is the change in the kinetic energy of the child plus the disk? (c) where has most of this kinetic energy gone? (d) Calculate the change in linear momentum of the system consisting of the child plus the disk (but not including the axle), from just before to just after impact. What caused this change in the linear momentum? (e) The child on the disk walks inward on the disk and ends up standing at a new location a distance from the axle. Now what is the angular speed? (f) What is the change in the kinetic energy of the child plus the disk, from the beginning to the end of the walk on the disk? (g) What was the source of this increased kinetic energy?

Short answer

The final angular velocity of the disk is $0.727rad/s$.

The final kinetic energy of the system is$-30.6J$ .

The magnitude of the linear momentum of the disk is$84.048kg.m/s$.

The final angular speed of the system is $1.313rad/s$.

The change in kinetic energy is $28.6J$.

Step-by-step solution

Definition of Kinetic energy and the linear momentum.

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.

Define the conservation of angular momentum.

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.

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

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

Find the angular velocity of the disk.

The final angular momentum of the system is

$L_f=I_f+{\omega }_f$

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

$$\begin{gathered} {\overrightarrow{\tau }}_{net}=\frac{d\overrightarrow{L}}{dt} \\ 0=\frac{d\overrightarrow{L}}{dt} \\ \overrightarrow{L_f}={\overrightarrow{L}}_i \end{gathered}$$

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

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

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

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

$=0.727rad/s$

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

Find the final kinetic energy of the system.

(b) Initial kinetic energy of the child is $K_C=\frac{1}{2}m{v}^2$

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,${\omega }_f=0.727rad/s$

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

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

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

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

(c) The loss in kinetic energy is used to do work against friction force.

Find the X component of linear momentum.

(d.) The magnitude of the linear momentum of the disk just after the collision is,

$$\begin{gathered} =p_f=(m+M) \\ =(25kg+43kg)(1.236m/s) \\ =84.048kg.m/s \end{gathered}$$

(e) Take the$x-axis$to be in the direction of the initial velocity of the child.

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

$$\begin{gathered} \Delta p_x=p_{f,x}=p_{i,x} \\ =84.048kg.m/s-57.5kg.m/s \\ =26.548kg.m/s \end{gathered}$$

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

Find the angular speed of the disk.

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.

$I_i{\omega }_i=I_f+{\omega }_f$

The final angular speed of the system can be calculated as

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

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

Find the change in Kinetic energy.

(f) The expression for final rotational kinetic energy of the child-disk system is,

$k_f=\frac{1}{2}{I}_f{{\omega }_f}^2$

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

$k_f=\frac{1}{2}{I}_f{{\omega }_f}^2.$

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

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$28.6J$

(g.) The work-energy theorem states that the work done by the system is equal to the change in kinetic energy. The work is done by the child, while he is walking inwards on the disk. Therefore, the change in kinetic energy is due to the work done by the child.