
What torque will increase the angular velocity of a solid disc of mass $16kg$ and diameter $1m$ from $0{\text{ to 2rpm}}$ in$8{\text{ }}\sec ond$?
$(A){\text{ }}\dfrac{\pi }{4}N - m$
$(B){\text{ }}\dfrac{\pi }{2}N - m$
$(C){\text{ }}\dfrac{\pi }{3}N - m$
$(D){\text{ }}\pi N - m$
Answer
232.8k+ 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
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

