Question

Suppose the bob of the previous problem has a speed of 1.4 m/s when the string makes an angle of 0.20 radian with the vertical. Find the tension at this instant. You can use cos θ ≈ 1 − θ2/2 and SINθ ≈ θ for small θ.

Answer 1

The tension at the instant is 1.176 N

Explanation:

As we know that,

$\begin{equation}T=m g \cos \theta+\frac{m v^{2}}{r}$  

Given in previous question  

Mass (m) = 100 g  

Converting gram to kilogram  

$\begin{equation}\frac{100}{1000}=0.1 \mathrm{kg}$

Mass = 0.1 kg

Speed ( v ) = 1.4 m/s

The length of bob of a simple pendulum  is 1 m

r= 1 m

The string makes an angle (θ) of almost 0.20 radian with the vertical

$\begin{equation}\theta=0.2$

g = 10 $m/s^2$ (The gravity of Earth )

$\begin{equation}\begin{aligned}T=0.1 \times 10 \times \cos \theta+\frac{0.1 \times 1.4^{2}}{1} \\T=1 \times \cos \theta+\frac{0.1 \times 1.96}{1} \\T=\cos \theta+0.196\end{aligned}$

$\begin{equation}\text { use } \cos \theta \approx 1-\frac{\theta^{2}}{2}$

$\begin{equation}\begin{aligned}&T=\left(1-\frac{\theta^{2}}{2}\right)+0.196\\&T=\left(1-\frac{0.2^{2}}{2}\right)+0.196\\&T=\left(1-\frac{0.04}{2}\right)+0.196\\&T=(1-0.02)+0.196\\&T=0.98+0.196\\&T=0.98+0.196\\&T=1.176\end{aligned}$

T = 1.176 N

The tension at this instant is 1.176 N

Answer 2

Answer:

इट इज द प्रीवियस प्रॉब्लम है द स्पीड ऑफ स्ट्रिंग मिक्स एंड एंगल ऑफ वर्टिकल एंड इंस्टेंट प्रमोशन❄️❣️