In outer space a rock of mass 4kgis attached to a long spring and swung at constant speed in a circle of radius 9m. The spring exerts a force of constant magnitude 760N.

(a) What is the speed of the rock?

(b) What is the direction of the spring force?

(c)The relaxed length of the spring is 8.7m. What is the stiffness of this spring?

A rock of mass $m=4\mathrm{kg}$is attached to a long spring and swung at constant speed in a circle of radius $R=9\mathrm{m}$. The spring exerts a force of constant magnitudeand the relaxed length of spring is$L=8.7\mathrm{m}$

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

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 net force$F_{net}$on the object is given by

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

The rock's speed is affected by the parallel rate of change of momentum and given the speed is constant, therefore, the parallel rate is zero, and it equals the size of the momentum rate change.

$\frac{d\overrightarrow{p}}{dt}=0$

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

At speeds far slower than the speed of light, the net force exerted on the rock equals the rate change of momentum, and the magnitude of the perpendicular rate change is given by

$$\begin{gathered} F_{net}=\frac{d\overrightarrow{p}}{dt} \\ =\frac{m{v}^2}{R} \end{gathered}$$

Also, the rock is under two forces, the tension force $F_T$and its weight, so the net force of both forces is

$F_{net}=F_T+mg$

The weight force is ignored in outer space.

$F_{net}=F_T$

Therefore,

$$\begin{gathered} F_T=\frac{m{v}^2}{R} \\ v=\sqrt{\frac{F_{T}R}{m}} \end{gathered}$$

(Solve for $v$)

Now put the values for $F_T$, $R$and $m$to get the speed of the rock

$$\begin{gathered} v=\sqrt{\frac{F_{T}R}{m}} \\ =\sqrt{\frac{760\mathrm{N}\left(9\mathrm{m}\right)}{4\mathrm{kg}}} \\ =41.35\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the rock is $41.35\mathrm{m}/\mathrm{s}$.

The motion is circular, and the rock has the potential to repeat the period. As a result, the rock's motion and momentum are tangential, and the direction of change in momentum is toward the path's centre.

So, the force is directed toward the path's centre.

The tension force$F_T$in the spring,

$F_T=kx$

Where$k$ is the stiffness of the spring and$x$ is the stretched length or the elongation that occurs to the spring.

And it is equal the difference between the initial length and the final length of the spring after elongation and we could calculate it by

$$\begin{gathered} x=R-L \\ =9\mathrm{m}-8.7\mathrm{m} \\ =0.3\mathrm{m} \end{gathered}$$

Now put the values for$F_T$and$x$,

$$\begin{gathered} k=\frac{R-L}{x} \\ =\frac{760\mathrm{N}}{0.3\mathrm{m}} \\ =2533\mathrm{N}/\mathrm{m} \end{gathered}$$

Hence, the stiffness of the spring is$2533\mathrm{N}/\mathrm{m}$.