Answer
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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.
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