A child of mass 35kgsits on a wooden horse on a carousel. The wooden horse is 3.3mfrom the center of the carousel, which rotates at a constant rate and completes one revolution every 5.2s.

(a) What are the magnitude and direction (tangential in direction of velocity, tangential in the opposite direction of the velocity, radial outward, radial inward) of $\left(d\left|\overrightarrow{p}\right|/dt\right)\stackrel{\mathbf{⏜}}{\mathbf{p}}$, the parallel component of $\mathit{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathit{d}\mathit{t}$for the child?

(b) What are the magnitude and direction of $\mathbf{\backslash }\overrightarrow{\mathbf{p}}\mathbf{\backslash }\mathit{d}\overline{\mathbf{p}}\mathbf{/}\mathit{d}\mathit{t}$. the perpendicular component of $\mathit{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathit{d}\mathit{t}$for the child?

(c) What are the magnitude and direction of the met force acting on the child? (d) What objects in the surroundings contribute to this horizontal net force acting on the child? (There are also vertical forces, but these cancel each other if the horse doesn't move up and down.)

A child of mass m=35kg sits on a wooden horse on a carousel. The wooden horse is R=3.3m from the center of the carousel, which rotates at a constant rate and completes one revolution every t=5.2s.

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.

(a)

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two components.

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 object's speed changes as the parallel rate of change of momentum changes, and as we know, the speed is constant, therefore, the parallel rate is zero and it is equal to the rate change of the magnitude of the momentum,

$$\begin{gathered} \frac{d\overrightarrow{p}}{dt}=\frac{d\backslash \overrightarrow{p}\backslash }{dt}\stackrel{⏜}{p} \\ =0 \end{gathered}$$

(b)

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,

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

The speed is the change in the distance with time.

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

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

The circumference is given by,$2\mathrm{\pi R}$

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

So, the distance is,$d=2\mathrm{\pi R}$

Therefore,

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

Now plug the values for R and t to get the speed of the wooden horse,

localid="1656687976382" $V=\frac{2\mathrm{\pi R}}{t}$

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

_{$$\begin{gathered} \left|\overrightarrow{\mathrm{p}}\right|\frac{\mathrm{d}\stackrel{⏜}{\mathrm{p}}}{\mathrm{dt}}=\frac{{\mathrm{mv}}^2}{\mathrm{R}} \\ =\frac{\left(35\mathrm{kg}\right)\left(3.98\mathrm{m}/\mathrm{s}\right)}{3.3\mathrm{m}} \\ =168\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$}

(c)

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

$$\begin{gathered} F_{net}=\frac{d\overrightarrow{p}}{dt} \\ =\frac{d\overrightarrow{p}}{dt} \\ =168N \end{gathered}$$

(d)

Because 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.