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Physics Lesson 7.2.4 - Newton's Second Law for Rotational Motion

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Welcome to our Physics lesson on Newton's Second Law for Rotational Motion, this is the fourth lesson of our suite of physics lessons covering the topic of Dynamics of Rotational Motion, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Newton's Second Law for Rotational Motion

From the analogy used to study rotational quantities in terms of linear ones, it is easy to derive the Newton's Second Law for Rotational Motion. Thus, since moment of inertia I in rotational motion is analogue to mass m in linear one, and angular acceleration α is analogue to the linear acceleration a, and giving that the Newton's Second Law of Motion in Linear Dynamics is F = m × a, we can write the Newton's Second Law of Motion as

τ = I × α

where τ is the torque, which is analogue to force in linear dynamics.

Example 3

A 12 N force pushes perpendicularly a 4 m long metal rod as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Dynamics of Rotational Motion

If the mass of the rod is 1 kg, calculate

  1. Angular acceleration if the rod starts rotating from rest around its centre without any friction.
  2. The number of rotation the bar makes in 10 s.

Solution 3

a. First, we calculate the torque τ. Thus, since the bar rotates around its centre, the rotating arm is d = L / 2 = 4 m / 2 = 2 m and

τ = d × F
= 2 m × 12 N
= 24 N × m

Now, let's calculate the rod's moment of inertia I. We use the formula of the long bar rotating around its centre (case 1). Thus, we have

I = 1/12 m × L2
= 1/12 × 1 × 42
= 1.25 kg × m2

Now, applying the Newton's Second law for rotational motion

τ = I × α

we calculate the angular acceleration α (given that 1 N = 1 kg × m / s2):

α = τ/l
= 24 kg × m2/s2/1.25 kg × m2
= 19.2 s - 2
= 19.2 rad/s2

b. First let's calculate the angular displacement φ. Given that the bar starts rotating from rest, we have ω0 = 0. Therefore, applying the kinematic equation

φ = ω0 × t + α × t2/2

we obtain after substitutions (t = 10 s):

φ = 0 × 10 + 19.2 × 102/2
= 960 rad

Therefore, giving that 1 rot = 2π rad, we obtain for the number of rotations N:

N = φ/
= 960 rad/2 × 3.14 rad/rot
= 152.9 rotations

You have reached the end of Physics lesson 7.2.4 Newton's Second Law for Rotational Motion. There are 8 lessons in this physics tutorial covering Dynamics of Rotational Motion, you can access all the lessons from this tutorial below.

More Dynamics of Rotational Motion Lessons and Learning Resources

Rotation Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
7.2Dynamics of Rotational Motion
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
7.2.1Moment of Force as Analogue of Force in Translational World
7.2.2Moment of Inertia
7.2.3Moment of Inertia in Different Systems of Rotational Motion
7.2.4Newton's Second Law for Rotational Motion
7.2.5Angular Momentum
7.2.6Work in Rotational Motion
7.2.7Kinetic Energy in Rotational Motion
7.2.8Power in Rotational Motion

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  5. Check your calculations for Rotation questions with our excellent Rotation calculators which contain full equations and calculations clearly displayed line by line. See the Rotation Calculators by iCalculator™ below.
  6. Continuing learning rotation - read our next physics tutorial: Centripetal Force

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