Question

Certain neutron stars are believed to be rotating at about 1rev/s. If such a star has a radius of 20km, the acceleration of an object on the equator of the star will be.
A. $20\times {{10}^{8}}m/{{s}^{2}}$
B. $8\times {{10}^{5}}m/{{s}^{2}}$
C. \[120\times {{10}^{5}}\]
D. $4\times {{10}^{8}}m/{{s}^{2}}$

Answer 1

Hint: Since this question is talking about an object in rotational motion we’ll talk about the centripetal acceleration which is given by the formula: \[\alpha ={{\omega }^{2}}r\], where $\alpha $is the centripetal acceleration, $\omega $ is the angular velocity and $r$is the radius of the circular path.

Formula used: \[\alpha ={{\omega }^{2}}r\], where $\alpha $is the centripetal acceleration, $\omega $ is the angular velocity and $r$is the radius of the circular path.
$\omega =2\pi f$, where $f$ is the frequency of the object.

Complete Step by Step Solution:
As is well known, a body in uniform circular motion is affected by an acceleration that is directed along the radius towards the centre of the circular route. Centripetal acceleration is the name given to that acceleration.
When a particle moves in a circle, its rate of acceleration is given by,
\[\alpha ={{\omega }^{2}}r\].

According to the given question $r=20km$
We know that $1km=1000m$ therefore $r=20,000m$
Since we know that $\omega =2\pi f$ and $f=1rev/s$
Then, $\omega =2\pi (1)rad/s$
Putting all the values in \[\alpha ={{\omega }^{2}}r\], we get:
$\alpha ={{(2\pi )}^{2}}\times 20000m/{{s}^{2}}$
$=4{{\pi }^{2}}\times 20000$, Putting $\pi =3.14$
$=4\times {{(3.14)}^{2}}\times 20000$
On solving the above equation we get,
$\alpha =8\times {{10}^{5}}m/{{s}^{2}}$
Hence the answer is option B. $8\times {{10}^{5}}m/{{s}^{2}}$

Note: In order to understand uniform circular motion in this kind of problem, we must first recall certain fundamental concepts. Then, in order to solve the problem, we will make use of the relation between angular velocity and angular acceleration. We can easily locate both of them using this relationship when the frequency and radius of the circular path are known. Through this, we can acquire the needed solution.