Question

An object moves in a circle of radius R at constant speed with a period T. If you want to change only the period in order to cut the object's acceleration in half, the new period should be A) Th/2 B) 7V2. C) T/4. D) 4T E) TI2

Answer 1

Answer: Option (B) is the correct answer.

Explanation:

Expression for centripetal acceleration is as follows.

                a = $r \omega^{2}$ .......... (1)

Also, we know that

             $\omega = \frac{2 \pi}{T}$ ........... (2)

Putting the value from equation (2) into equation (1) as follows.

             a = $r (\frac{2 \pi}{T})^{2}$

                = $r \frac{2 (\pi)^{2}}{4}$

As,       $a \propto \frac{1}{T^{2}}$

               a = $\frac{k}{T^{2}}$

or,              $aT^{2} = k$

Now, we will reduce a to $\frac{a}{2}$. So, new value of $T^{2}$ will be equal to $2T^{2}$.

Therefore, value of new period will be as follows.

           $T' = \sqrt{2T^{2}}$

                      = $\sqrt{2}T$

Thus, we can conclude that the new period is equal to $T \sqrt{2}$.