A string of length 2 m can withstand a maxim | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The centripetal force required for circular motion is provided by the tension in the string.$T = mr\omega^2$Given:Mass $m = 2 \ kg$Radius $r = 2 \

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) The centripetal force required for circular motion is provided by the tension in the string.$T = mr\omega^2$Given:Mass $m = 2 \ kg$Radius $r = 2 \ m$Maximum tension $T_{\max} = 16000 \ N$Angular velocity $\omega = 2\pi f$,where $f$ is the frequency in rotations per second.Substituting the values into the formula:$T_{\max} = mr(2\pi f_{\max})^2$$16000 = 2 	imes 2 	imes 4\pi^2 f_{\max}^2$$16000 = 16\pi^2 f_{\max}^2$$f_{\max}^2 = \frac{16000}{16\pi^2} = \frac{1000}{\pi^2}$Using the approximation $\pi^2 \approx 10$:$f_{\max}^2 = \frac{1000}{10} = 100$$f_{\max} = \sqrt{100} = 10 \ 	ext{rotations per second}$.

Similar Questions

Find the centripetal force acting on a coin weighing $0.1 \,kg$ placed $0.1 \,m$ from the centre on a gramophone disc rotating at $600 \,rpm$.

$A$ cage revolves around a vertical circle of radius $R$ with constant linear speed $\sqrt{gR}$. The cage is connected to the revolving arm in such a manner that a boy of mass $m$ remains always vertical while standing on a weighing machine kept inside the cage. It is found that

$A$ string of length $0.1\,m$ cannot bear a tension more than $100\,N$. It is tied to a body of mass $100\,g$ and rotated in a horizontal circle. The maximum angular velocity can be .......... $rad/s$.

$A$ train is moving towards north. At one place it turns towards north-east. Here,we observe that:

$A$ conical pendulum is moving in a circle with angular velocity $\omega$ as shown. If the tension in the string is $T$,which of the following equations is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module