Solveeit Logo
Login

Question

Question: A disc is rotating with an angular velocity \({\omega _0}\). A constant retarding torque is applied ...

A disc is rotating with an angular velocity ω0{\omega _0}. A constant retarding torque is applied on it to stop the disc. The angular velocity becomes ω02\dfrac{{{\omega _0}}}{2} after nn rotations. How many more rotations will it make before coming to rest?
A. nn
B. 2n2n
C. n2\dfrac{n}{2}
D. n3\dfrac{n}{3}

Explanation

Solution

In order to solve this question, we should know that a torque produce angular acceleration in the body and since torque is constant then, angular acceleration will also be constant but in negative direction as its retarding torque, here we will use general equation of motion to calculate number of turns a disc will make before it came at rest.

Formula used:
If ω,ω0,θ,α\omega ,{\omega _0},\theta ,\alpha be the final angular velocity, initial angular velocity, displacement, angular acceleration then,
ω2−ω02=2αθ{\omega ^2} - {\omega _0}^2 = 2\alpha \theta

Complete step by step answer:
According to the question, we have torque is retarding so, angular acceleration produced by this torque will also be retarding which will be −α- \alpha and for first n rotations the velocities have,
final velocity ω=ω02\omega = \dfrac{{{\omega _0}}}{2}
initial velocity is ω0{\omega _0}
angular acceleration −α- \alpha
Displacement is n rotations so using,
ω2−ω02=2αθ{\omega ^2} - {\omega _0}^2 = 2\alpha \theta we get,
⇒ω024−ω02=−2αn\Rightarrow \dfrac{{{\omega _0}^2}}{4} - {\omega _0}^2 = - 2\alpha n
⇒34ω02=2αn→(i)\Rightarrow \dfrac{3}{4}{\omega _0}^2 = 2\alpha n \to (i)

Now let n’ be the further rotations by disc to came at rest at conditions,
final velocity ω=0\omega = 0
initial velocity ωinitial=ω02{\omega _{initial}} = \dfrac{{{\omega _0}}}{2}
angular acceleration −α- \alpha
displacement is n’ rotations so using,
ω2−ω02=2αθ{\omega ^2} - {\omega _0}^2 = 2\alpha \theta we get,
0−ω024=−2αn′0 - \dfrac{{{\omega _0}^2}}{4} = - 2\alpha n'
⇒ω024=2αn′→(ii)\Rightarrow \dfrac{{{\omega _0}^2}}{4} = 2\alpha n' \to (ii)
Now, divide first equation by second we get,
34ω02ω024=2αn2αn′\dfrac{{\dfrac{3}{4}{\omega _0}^2}}{{\dfrac{{{\omega _0}^2}}{4}}} = \dfrac{{2\alpha n}}{{2\alpha n'}}
⇒nn′=3\Rightarrow \dfrac{n}{{n'}} = 3
∴n′=n3\therefore n' = \dfrac{n}{3}

Hence, the correct option is D.

Note: It should be remembered that, retarding torque means its applied in such a ways that it opposes the rotational motion of the body and thus produced negative angular acceleration on the body and always check initial and final velocity of the bodies while using equations of motion to solve such problems.