Question

In the dark in outer space, you observe a glowing ball of known mass 2kgmoving in the xyplane at constant speed in a circle of radius, 6.5 m with the centre of the circle at the origin$\left(\left(0,0,0\right)m\right)$. You can't see what's making it move in a circle. At time t=0 the ball is at location$\mathbf{(}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{m}$and has velocity$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{40}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{m}\mathbf{/}\mathbf{s}$.

On your own paper draw a diagram of the situation showing. the circle and showing the position and velocity of the ball at time $\mathit{r}\mathbf{=}\mathbf{0}$. The diagram will help you analyse the situation. Use letters a-j figure 5.75) to answer questions about directions ( +xto the right, +yup).

At time:$\mathbf{t}\mathbf{=}\mathbf{0}$

(a) What is the direction of the vector$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}$?

(b) What are the magnitude and direction localid="1656743973413" $\mathbf{\left(}\mathbf{d}\left|\stackrel{\rightharpoonup }{p}\right|\mathbf{d}\mathbf{t}\mathbf{\right)}\mathit{\rho }$ofthe parallel component of$\mathit{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}\mathbf{/}\mathit{d}\mathit{t}$?

(c) What are the magnitude and direction oflocalid="1656744314609" $\mathbf{\left|}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}\mathbf{\right|}\mathit{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}\mathbf{/}\mathit{d}\mathit{t}$, the perpendicular component of$\mathit{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}\mathbf{/}\mathit{d}\mathit{t}$?

(d) Even though you can't see what's causing the motion, what can you conclude must be the direction of the vector${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}}_{\mathbf{n}\mathbf{e}\mathbf{t}}$?

(e) Even though you can't see what's causing the motion, what can you conclude must be the vector${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}}_{\mathbf{n}\mathbf{e}\mathbf{t}}$?

(f) You learn that at time, two forces act on the ball, and that at this instant one of these forces is${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}}_{\mathbf{1}}\mathbf{=}\left\{196,-369,0\right\}\mathit{N}$. What is the other force?

Short answer

a. The direction of the vector $\stackrel{\rightharpoonup }{p}p$is a.

b. The magnitude is 0 and its direction is j.

c. The magnitude is $492kg.m/s^2$and its direction .

d. The direction of ${\stackrel{\rightharpoonup }{F}}_{net}$is, C.

e. The vector of role="math" localid="1656739509533" ${\stackrel{\rightharpoonup }{F}}_{net}$ is .

f. The other force is .

Step-by-step solution

Given data

The ball of mass 2kg moving in the xy plane at constant speed in a circle of radius 6.5m, with the centre of the circle at the origin $\left(\left(0,0,0\right)m\right)$. You can't see what's making it move in a circle. At time t=0 the ball is at location and has velocity$\left(0,40,0\right)m/s$.

Definition of force.

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

Find the direction of the vector p⇀.

(a)

The ball is at x = 6.5 in the negative direction of the x - axis.

The diagram is shown in the figure below where the speed is in the positive direction of the y -axis.

The direction of the momentum will be in the positive direction of the y -axis and from figure 5.75 , the direction which represents the +y -direction is.

As the velocity is given by , therefore, the momentum $\stackrel{\rightharpoonup }{p}$is in +y -direction.

Find magnitude and direction of (d|p⇀|dt)ρ the parallel component of dp⇀/dt .

(b)

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 two parts that we are taken about are the parallel rate of change of the momentum$\frac{d\stackrel{\rightharpoonup }{p}}{dt}$ and the perpendicular rate of change of the momentum$\frac{d\stackrel{\rightharpoonup }{p}}{dt}$.

So, the change of momentum required,

$\frac{d\stackrel{\rightharpoonup }{p}}{dt}=\frac{d\stackrel{\rightharpoonup }{p}}{dt}+\frac{d\stackrel{\rightharpoonup }{p}}{dt}$

The parallel rate of change of the momentum changes the speed of the object and as we are given the speed is constant, therefore, the parallel rate is zero and it is equal to the rate change of the magnitude of the momentum

$$\begin{gathered} \frac{d\stackrel{\rightharpoonup }{p}}{dt}=\frac{d\left|\stackrel{\rightharpoonup }{p}\right|}{dt}\hat{p} \\ =0 \end{gathered}$$

So, the direction the rate change of the magnitude of the momentum is zero and it represents by direction from figure 5.75 .

Find magnitude and direction of , |p⇀|dp⇀/dtthe perpendicular component of dp⇀/dt .

(c)

The shift in direction caused by the perpendicular rate of change of momentum is referred to as the rate 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

$$\begin{gathered} \frac{d\hat{p}}{dt}=\left|\stackrel{\rightharpoonup }{P}\right|\frac{d\hat{p}}{dt} \\ =\frac{m{v}^2}{R} \end{gathered}$$

The magnitude of speed for the ball,

$$\begin{gathered} v=\sqrt{x^2+y^2+z^2} \\ =\sqrt{0^2+{40}^2+0^2} \\ =40m/s \end{gathered}$$

Now put the values for $\mathcal{m}$, $\nu$and R to get $\left|\stackrel{\rightharpoonup }{P}\right|\frac{d\hat{p}}{dt}$

role="math" localid="1656908330163" $$\begin{gathered} \left|\stackrel{\rightharpoonup }{P}\right|\frac{d\hat{p}}{dt}=\frac{m{v}^2}{R} \\ =\frac{\left(2kg\right){\left(40m/s\right)}^2}{6.5m} \\ =492kg.m/s^2 \end{gathered}$$

The direction changes in the +y -direction, from figure 5.75 , this direction which represents the +y -direction is .

Find the direction of F⇀net .

(d)

The ball's motion is circular, and it has the potential to repeat the period.

As a result, the ball's motion and momentum are tangential, and the direction of change in momentum is toward the circle's centre.

Hence, the direction of the net force is toward the centre of the circle in the positive direction of the X - axis as shown in the figure in part (a) which represents in figure 5.75 by, C .

Find the vector of F⇀net .

(e)

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

$$\begin{gathered} {\stackrel{\rightharpoonup }{F}}_{net}=\frac{d\stackrel{\rightharpoonup }{p}}{dt} \\ =\left|\stackrel{\rightharpoonup }{p}\right|\frac{d\hat{p}}{dt} \\ =492N \end{gathered}$$

As this change in direction is in +y direction, hence the force is in one direction.

The y component of the force$F_y=492N$.

Also, there is no force component in z-direction, therefore,$F_z=0$.

The force vector in the three directions is given by

$\stackrel{\rightharpoonup }{F}=F_x\hat{i}+F_y\hat{j}+F_z\hat{k}$

Now put the values for localid="1656910231380" $F_{x,}{F}_{y}and{F}_z$) to get the vector of the tension force

Find the other force.

(f)

We are given two forces act on the ball $F_1$and $F_2$at time t = 0 when the net force is .

Where

The sum of two vectors can be expressed as any vector, so, write the exerted force ${\stackrel{\rightharpoonup }{F}}_{net}$as the sum of two other forces: ${\stackrel{\rightharpoonup }{F}}_1$and ${\stackrel{\rightharpoonup }{F}}_2$

$$\begin{gathered} {\stackrel{\rightharpoonup }{F}}_{net}={\stackrel{\rightharpoonup }{F}}_1+{\stackrel{\rightharpoonup }{F}}_2 \\ {\stackrel{\rightharpoonup }{F}}_2={\stackrel{\rightharpoonup }{F}}_{net}-{\stackrel{\rightharpoonup }{F}}_1 \end{gathered}$$

Now put the values for ${\stackrel{\rightharpoonup }{F}}_{net}$ and ${\stackrel{\rightharpoonup }{F}}_1$ to get ${\stackrel{\rightharpoonup }{F}}_2$

The other force is