Certain neutron stars are believed to be rota | vedclass

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

Question

(B) The acceleration of an object on the equator of a rotating star is the centripetal acceleration,given by $a = \omega^2 r$.Given frequency $n = 1 \

Vedclass · Accepted Answer

$8 	imes 10^5 \, m/sec^2$

Vedclass · Answer

(B) The acceleration of an object on the equator of a rotating star is the centripetal acceleration,given by $a = \omega^2 r$.Given frequency $n = 1 \, rev/sec$,so angular velocity $\omega = 2\pi n = 2\pi \, rad/sec$.The radius $r = 20 \, km = 20 	imes 10^3 \, m$.Substituting these values into the formula:$a = (2\pi 	imes 1)^2 	imes (20 	imes 10^3)$$a = 4\pi^2 	imes 20 	imes 10^3$Using $\pi^2 \approx 10$:$a \approx 4 	imes 10 	imes 20 	imes 10^3 = 800 	imes 10^3 = 8 	imes 10^5 \, m/sec^2$.

Certain neutron stars are believed to be rotating at about $1 \, rev/sec$. If such a star has a radius of $20 \, km$,the acceleration of an object on the equator of the star will be:

Similar Questions

$A$ particle moves along a circular path of radius $r = \frac{20}{\pi} \ m$ with a constant tangential acceleration. If the velocity of the particle at the end of the second revolution is $80 \ m/s$,what is the tangential acceleration?

$A$ particle performing uniform circular motion of radius $\frac{\pi}{2} \ m$ makes $x$ revolutions in time $t$. Its tangential velocity is

What is the value of the tangential component of the linear acceleration of a particle of a rigid body rotating with a constant angular velocity?

The angular acceleration of a body, moving along the circumference of a circle, is :

$A$ particle starting from rest moves in a circle of radius $r$. It attains a velocity of $V_{0} \; m/s$ in the $n^{\text{th}}$ round. Its angular acceleration will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module