Question

A person stands on a spring balance at the equator.
(a) By what friction is the dinbalance reag less than hits true weight? (b) If the speed of earth's rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case?

Answer 1

(a) Friction = 3.45 X $10^{-3}$

(b) The length of the day will be 2 hours

Explanation:

(a)

Step 1:

We know the Formula

$\begin{equation}\begin{aligned}&g^{\prime}=g-\omega^{2} r\\&g-g^{\prime}=\omega^{2} r\end{aligned}$

Divide both side by g

$\begin{equation}\frac{g-g^{\prime}}{g}=\frac{\omega^{2} r}{g}$

$\begin{equation}\frac{\underline{s}-\underline{g}^{\prime}}{g}\en$ is the fraction of weight which is less than its true weight

Step 2:

Fraction of weight is given by = $\begin{equation}\frac{\omega^{2} r}{g}=3.45 \times 10^{-3}$

(b). For this situation the outward power because of pivot of earth= Half the genuine weight

$\begin{equation}\begin{aligned}&m w^{2} R=\frac{1}{2} m g\\&m w^{2} R=\frac{1}{2} m g\\&2\left(\frac{2 \pi}{T}\right)^{2} R=g\\&8 \pi^{2} R=g T^{2}\\&T^{2}=\frac{8 \pi^{2} R}{g}\end{aligned}$

Here

π = 3.14  

Radius ( R )  = 6400 km  

Converting km to m  

Radius ( R )  = 6400×1000

g = 9.8 m/s2  (The gravity of Earth)

On substituting the values,

$\begin{equation}\begin{aligned}&T^{2}=\frac{8 \times 3.14^{2} \times 6400 \times 1000}{9.8}\\&T^{2}=\frac{8 \times 9.85 \times 6400000}{9.8}\\&\begin{aligned}T^{2} &=\frac{78.87 \times 6400000}{9.8} \\T^{2} &=\frac{504811520}{9.8} \\T^{2} &=51511379.59\end{aligned}\end{aligned}\end$

Square root both side

$\begin{equation}\begin{array}{c}{\sqrt{T^{2}}=\sqrt{51511379.59}} \\{T=7177.14 \sec }\end{array}\end$

Converting Second to hour - divide by 3600

$\begin{equation}T=\frac{7177.14}{3600} h r$

$\begin{equation}T=1.993 h r \approx 2 \text {hour}$

Answer 2

Answer:

द फ्रिक्शन इस द डिसबैलेंस एक्शन लिस्ट एंड रूबी रेड द स्पीड ऑफ अर्थ रोटेशन रिसर्च अमाउंट