A child of mass 40kgsits on a wooden horse on a carousel. The wooden horse is 5mfrom the center of the carousel, which completes one revolution every 90s. What is$\left(d\overrightarrow{p}/\mathrm{dt}\right)\stackrel{\mathbf{⏜}}{\mathbf{p}}$for the child, both magnitude and direction? What is$\left|\overrightarrow{p}\right|\left(d\stackrel{⏜}{p}/\mathrm{dt}\right)$for the child? What is the net force acting on the child? What objects in the surroundings exert this force?

A child of mass m=40kg sits on a wooden horse on a carousel. The wooden horse is R=5m from the center of the carousel, which completes one revolution every t=90s .

A force is a push or pull on an object caused by the interaction of the thing with another object. Every time two things interact, a force is exerted on each of them. The two items no longer feel the force after the interaction ends.

The change in distance over time is the speed.

$\mathbf{v}\mathbf{=}\frac{\mathbf{d}}{\mathbf{t}}$

The Earth moves above the circumference of a circle when it completes one round.

The circumference,

$\mathbf{2}\mathbf{\pi R}$

Where Ris the radius of the circle (distance between the wooden horse and the center of the carousel).

So, the distance,

$\mathit{d}\mathbf{=}\mathbf{2}\mathbf{\pi R}$

Substitutein speed’s equation.

$\mathrm{v}=\frac{2\mathrm{\pi R}}{\mathrm{t}}$

Now put the values for R and t,

$$\begin{gathered} v=\frac{2\mathrm{\pi R}}{t} \\ =\frac{2\mathrm{\pi }\left(5\mathrm{m}\right)}{90\mathrm{s}} \\ =0.34\mathrm{m}/\mathrm{s} \end{gathered}$$

So, the speed of the wooden horse is 0.34m/s.

The parallel rate of change of momentum $\frac{d\overrightarrow{p}}{dt}$and the perpendicular rate of change of momentum $\frac{d\overrightarrow{p}}{dt}$are the two elements that we are concerned with.

So, the change of momentum required is given by,

$\frac{d\overrightarrow{p}}{dt}=\frac{d\overrightarrow{p}}{dt}+\frac{d\overrightarrow{p}}{dt}$

The parallel rate of change of momentum affects the object's speed, and since the speed is constant, the parallel rate is zero and equal to the rate of change of the size of the momentum.

$\frac{d\overrightarrow{p}}{dt}=\frac{d\left|\overrightarrow{p}\right|}{dt}\stackrel{⏜}{p}$

=0

As a result, the rate change is the direction change owing to the perpendicular rate of change.

The magnitude of the perpendicular rate change equals the rate change of the direction of the momentum and at speeds much less than the speed of light is given by

$$\begin{gathered} \frac{d\overrightarrow{p}}{dt}=\backslash \overrightarrow{p}\backslash \frac{d\stackrel{⏜}{p}}{dt} \\ =\frac{m{v}^2}{R} \end{gathered}$$

Now put the values for m ,v and R to get $\backslash \overrightarrow{p}\backslash \frac{d\stackrel{⏜}{p}}{dt}$

$$\begin{gathered} \frac{m{v}^2}{R}=\backslash \overrightarrow{P}\backslash \frac{d\stackrel{⏜}{p}}{dt} \\ =\frac{\left(40kg\right){\left(0.34m/s\right)}^2}{5m} \\ =0.97\mathrm{N} \end{gathered}$$

Hence, the magnitude of $\backslash \overrightarrow{P}\backslash \frac{d\stackrel{⏜}{p}}{dt}$is 0.97N and directed towards the center of the carousel.

The net force exerted on the object equals the rate change of the momentum and the magnitude of the perpendicular rate change at speeds much less than the speed of light is given by

$$\begin{gathered} F_{net}=\frac{d\overrightarrow{p}}{dt} \\ =\frac{d\overrightarrow{p}}{dt} \\ =0.97\mathrm{N} \end{gathered}$$

Therefore the net force is 0.97 N .

Since the wooden horse does not move up or down, no external forces contribute to the horizontal net force, and the vertical forces cancel each other out.