Answer
64.8k+ views
Hint: Torque is the product of the moment of inertia and the angular and the angular acceleration. We will calculate torque for the solid disk around the center. And for this, we have to first calculate the inertia for the center of the solid disk.
Formula used
Torque,
$ \Rightarrow \tau = I\alpha $; Where $\tau $is the torque, $I$ is the inertia, and $\alpha $is the angular acceleration.
Moment of Inertia for the center of the solid disk,
$ \Rightarrow I = \dfrac{1}{2}M{R^2}$; Where $M$ is mass of the object, $R$ is the radius.
Angular acceleration,
$ \Rightarrow \alpha = \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$ ; Where $n$ is the change in the rpm and $t$ is the time taken.
Complete step by step answer:
Since it is given in the question that there is a circular disk having mass 16kg and radius 1m. So we have to calculate the torque required for this.
As we know
$ \Rightarrow \tau = I\alpha $
And for calculating torque we have to first calculate the Moment of Inertia and angular acceleration.
For this we will use the below two formulas:
$ \Rightarrow I = \dfrac{1}{2}M{R^2}$
And
$ \Rightarrow \alpha = \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$
Combining and putting the equations in one
$ \Rightarrow \tau = I\alpha $
And we already know the values of these. So putting those values and we will get the required torque.
$ \Rightarrow \dfrac{1}{2}M{R^2} \times \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$
After substituting the values,
$ \Rightarrow 16{\left( {\dfrac{1}{2}} \right)^2} \times \dfrac{{\pi \left( {2 - 0} \right)}}{8}$
We get,
$ \Rightarrow \pi N - m$
So the torque required to increase the angular velocity of a solid circular disk will be\[\pi N - m\].
Therefore the Option D will be the right choice for this question.
Note: So when a force is applied to an object it begins to rotate with an acceleration reciprocally proportional to its moment of inertia. This relation is thought of like Newton’s Second Law for rotation. The moment of inertia is the rotational mass, and therefore, the force is rotational. Angular motion obeys Newton’s 1st Law.
Formula used
Torque,
$ \Rightarrow \tau = I\alpha $; Where $\tau $is the torque, $I$ is the inertia, and $\alpha $is the angular acceleration.
Moment of Inertia for the center of the solid disk,
$ \Rightarrow I = \dfrac{1}{2}M{R^2}$; Where $M$ is mass of the object, $R$ is the radius.
Angular acceleration,
$ \Rightarrow \alpha = \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$ ; Where $n$ is the change in the rpm and $t$ is the time taken.
Complete step by step answer:
Since it is given in the question that there is a circular disk having mass 16kg and radius 1m. So we have to calculate the torque required for this.
As we know
$ \Rightarrow \tau = I\alpha $
And for calculating torque we have to first calculate the Moment of Inertia and angular acceleration.
For this we will use the below two formulas:
$ \Rightarrow I = \dfrac{1}{2}M{R^2}$
And
$ \Rightarrow \alpha = \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$
Combining and putting the equations in one
$ \Rightarrow \tau = I\alpha $
And we already know the values of these. So putting those values and we will get the required torque.
$ \Rightarrow \dfrac{1}{2}M{R^2} \times \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$
After substituting the values,
$ \Rightarrow 16{\left( {\dfrac{1}{2}} \right)^2} \times \dfrac{{\pi \left( {2 - 0} \right)}}{8}$
We get,
$ \Rightarrow \pi N - m$
So the torque required to increase the angular velocity of a solid circular disk will be\[\pi N - m\].
Therefore the Option D will be the right choice for this question.
Note: So when a force is applied to an object it begins to rotate with an acceleration reciprocally proportional to its moment of inertia. This relation is thought of like Newton’s Second Law for rotation. The moment of inertia is the rotational mass, and therefore, the force is rotational. Angular motion obeys Newton’s 1st Law.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)