Question

An insect moving in a circle travels N1 revolutions in anticlockwise sense for a time T1 and N2 revolutions in
clockwise sense for a time T2.
Find the angular speed averaged over the time T1+ T2​

Answer 1

Given:

An insect moving in a circle travels N1 revolutions in anticlockwise sense for a time T1 and N2 revolutions in clockwise sense for a time T2.

To find:

Average angular speed.

Calculation:

Average angular speed can be defined as the total angular distance divided by the total time taken to cover that angular distance.

2π needs to be multiplied to the revolution factor in order to convert it into radians unit.

Average angular speed be $\omega$.

$\rm{ \omega = \dfrac{total \: angular \: distance}{total \: time} }$

$= > \rm{ \omega = \dfrac{(N1 \times 2\pi) + (N2\times 2\pi)}{T1 + T2} }$

$= > \rm{ \omega = \dfrac{(N1 + N2)\times 2\pi}{T1 + T2} }$

$= > \rm{ \omega = \dfrac{2(N1 + N2)\pi}{T1 + T2} }$

So, final answer is:

$\boxed{\bf{ \omega = \dfrac{2(N1 + N2)\pi}{T1 + T2} }}$

Answer 2

Explanation:

Angular displacement = $\sf θ=θ _1 + θ _2$

• $\sf θ _1 = 2πN_1$
• $\sf θ _2 = 2πN_2$

Angular speed :-

$\sf ω= \dfrac{θ}{t}$

$\sf ω _{av}= \dfrac{2π}{T _1 + T _2} \left| N_1 - N_2 \right|$