Question

The speed of revolution of particle moving around a circle is doubled and its angular speed is halved, then its centripetal acceleration

• A. remains unchanged
• B. halved
• C. doubled
• D. becomes four times

Answer 1

The correct option is A remains unchanged

Centripetal acceleration $a_r=r{w}^2$

Using $v=rw$ $⟹a_r=vw$

Now the speed is doubled but the angular speed is halved i.e $v^{\prime }=2v$ and $w^{\prime }=\frac{w}{2}$

$\therefore$ New centripetal acceleration $a_r^{\prime }=v^{\prime }{w}^{\prime }=\left(2v\right)\times \frac{w}{2}=vw=a_r$

Thus the centripetal acceleration remains the same.