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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.
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
where τ is the torque, which is analogue to force in linear dynamics.
A 12 N force pushes perpendicularly a 4 m long metal rod as shown in the figure.
If the mass of the rod is 1 kg, calculate
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
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
Now, applying the Newton's Second law for rotational motion
we calculate the angular acceleration α (given that 1 N = 1 kg × m / 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
we obtain after substitutions (t = 10 s):
Therefore, giving that 1 rot = 2π rad, we obtain for the number of rotations N:
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.
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