91Ó°ÊÓ

An unknown mass is attached to a spring. In outer space, far from other objects, you hold the other end of the spring and swing the ball around in a circle of radius at constant speed.

The magnitude of an object's rate of change of position with time, or the magnitude of change of position per unit of time, is its speed; it is thus a scalar quantity. The magnitude of the velocity is called speed. The product of mass and velocity is called momentum. This means that speed is a scalar quantity while momentum is a vector quantity

When the time for 10 round is 6.88 s, then for one round the time will be:

$$\begin{gathered} \mathrm{t}=\frac{6.88\mathrm{s}/\mathrm{round}}{10\mathrm{round}} \\ =0.688\mathrm{s} \end{gathered}$$

The change in distance over time is the speed. As a result, it is provided by

$v=\frac{d}{t}$ …………………… (1)

The ball moves above the diameter of a circle as it travels through one round. The circumference is calculated using

$2\mathrm{Ï€¸é}$

Where is the circle's radius. As a result, the ball's travel distance is

$\mathrm{d}=2\mathrm{Ï€¸é}$

To get the new form, enter the expression into equation (1).

$\mathrm{v}=\frac{2\mathrm{Ï€¸é}}{\mathrm{t}}$

Now we can insert our and numbers into equation (2) to calculate the engineer's speed.

$$\begin{gathered} \mathrm{v}=\frac{2\mathrm{Ï€¸é}}{\mathrm{t}} \\ =\frac{2\mathrm{Ï€}(1.5\mathrm{m})}{0.688\mathrm{s}} \\ =13.70\mathrm{m}/\mathrm{s} \end{gathered}$$

The speed of the ball is$13.70\mathrm{m}/\mathrm{s}$

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two parts. The parallel rate of change of momentum $\frac{\mathrm{d}\stackrel{→}{\mathrm{p}}}{{\mathrm{dt}}_{\left|\right|}}$and the perpendicular rate of change of momentum $\frac{\mathrm{d}\stackrel{→}{\mathrm{p}}}{{\mathrm{dt}}_{âŠ\yen }}$are the two elements that we are concerned with. As a result, the net force on the object is provided by ${\mathrm{F}}_{\mathrm{net}}$.

$$\begin{gathered} F_{net}=\frac{\mathrm{d}\stackrel{→}{\mathrm{p}}}{\mathrm{dt}} \\ =\frac{\mathrm{d}\stackrel{→}{\mathrm{p}}}{{\mathrm{dt}}_{\left|\right|}}+\frac{\mathrm{d}\stackrel{→}{\mathrm{p}}}{{\mathrm{dt}}_{âŠ\yen }} \end{gathered}$$

The parallel rate of change of momentum affects the ball'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.

$\frac{\mathrm{d}\stackrel{→}{\mathrm{p}}}{{\mathrm{dt}}_{\left|\right|}}=0$

As a result, the rate change is the direction change due to the perpendicular rate of change. At speeds much slower than the speed of light, the net force applied on the ball equals the rate change of momentum, and the magnitude of the perpendicular rate change is given by

$F_{net}=\frac{d\stackrel{→}{p}}{d{t}_{âŠ\yen }}$

The momentum's magnitude does not change, but its direction does.

The perpendicular force applied to the ball causes the momentum to change direction.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}={\mathrm{F}}_{âŠ\yen } \\ =\frac{\mathrm{d}\stackrel{→}{\mathrm{p}}}{{\mathrm{dt}}_{âŠ\yen }} \end{gathered}$$

The perpendicular force applied to the ball causes the momentum to change direction.

The spring's tension force ${\mathrm{F}}_{\mathrm{T}}$is given by,

${\mathrm{F}}_{\mathrm{T}}=\mathrm{kx}$ ……………………. (2)

Where represents the spring's stiffness (Force per unit length) and is 1000 N/m, and represents the cord's stretched length or elongation. And it's equal to the difference between the spring's starting length (relaxed length) L and its ultimate length after R elongation, which we can figure out by,

$$\begin{gathered} \mathrm{x}=\mathrm{R}-\mathrm{L} \\ =1.5\mathrm{m}-1.2\mathrm{m} \\ =0.3\mathrm{m} \end{gathered}$$

To solve the equation (2) for k by plugging in our and values.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{T}}=\mathrm{kx} \\ =(1000\mathrm{N}/\mathrm{m})(0.3\mathrm{m}) \\ =300\mathrm{N} \end{gathered}$$

The magnitude of the force is 300 N.

The centrifugal force exerted on the ball is caused by the tension force calculated in component (d), where the rate change is the change in direction owing to the perpendicular rate of change. At speeds far slower than the speed of light, the amount of the perpendicular rate change matches the rate change of the direction of the momentum.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{T}}=\frac{\mathrm{d}\stackrel{→}{\mathrm{p}}}{\mathrm{dt}} \\ =\frac{{\mathrm{mv}}^2}{\mathrm{R}} \end{gathered}$$ …………………….. (3)

Where denotes the ball's mass. To get , we can solve equation (3) for and plug in the values for ${\mathrm{F}}_{\mathrm{T}}$, v, and R.

$$\begin{gathered} \mathrm{m}=\frac{{\mathrm{RF}}_{\mathrm{T}}}{{\mathrm{v}}^2} \\ =\frac{(1.5\mathrm{m})\left(3000\mathrm{N}\right)}{{(13.70\mathrm{m}/\mathrm{s})}^2} \\ =2.40\mathrm{kg} \end{gathered}$$

The mass of the ball is 2.40Kg.