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Chapter 8: Problem 57

Derive the rotational form of Newton's second law as follows. Consider a rigid object that consists of a large number \(N\) of particles. Let \(F_{i}, m_{i},\) and \(r_{i}\) represent the tangential component of the net force acting on the ith particle, the mass of that particle, and the particle's distance from the axis of rotation, respectively. (a) Use Newton's second law to find \(a_{i}\), the particle's tangential acceleration. (b) Find the torque acting on this particle. (c) Replace \(a_{i}\) with an equivalent expression in terms of the angular acceleration \(\alpha\) (d) Sum the torques due to all the particles and show that $$\sum_{i=1}^{N} \tau_{i}=I \alpha$$

Short Answer

Expert verified
Answer: The rotational form of Newton's second law states that the net torque acting on a rigid object is equal to the product of its moment of inertia and angular acceleration, represented as: $$\sum_{i=1}^{N} \tau_{i} = I \alpha$$

Step by step solution

01

Find the tangential acceleration \(a_{i}\) using Newton's second law

According to Newton's second law, the net force acting on an object is equal to the product of its mass and acceleration. So, for the ith particle, we have: $$F_{i} = m_{i}a_{i}$$ Here, \(F_{i}\) is the tangential component of the net force, \(m_{i}\) is the mass, and \(a_{i}\) is the tangential acceleration of the ith particle. Now, we have expressed the tangential acceleration in terms of force and mass.
02

Find the torque acting on the ith particle

Torque is defined as the product of the force acting on an object and the distance of that force from the axis of rotation. For the ith particle, the torque \(\tau_{i}\) is given by: $$\tau_{i} = F_{i} r_{i}$$ where \(F_{i}\) is the tangential component of the net force acting on the ith particle and \(r_{i}\) is the distance of the ith particle from the axis of rotation.
03

Replace \(a_{i}\) with an equivalent expression in terms of angular acceleration \(\alpha\)

The tangential acceleration \(a_{i}\) is related to the angular acceleration \(\alpha\) as follows: $$a_{i} = r_{i} \alpha$$ Now, we can substitute this expression for \(a_{i}\) in the equation for torque in Step 2: $$\tau_{i} = F_{i} r_{i}$$ Substituting \(a_{i} = r_{i} \alpha\) and recalling that \(F_{i} = m_{i} a_{i}\), we get: $$\tau_{i} = m_{i} r_{i}^{2} \alpha$$
04

Sum the torques due to all the particles and show that the rotational form of Newton's second law holds

Now, we will sum the torques for all the particles in the rigid object: $$\sum_{i=1}^{N} \tau_{i} = \sum_{i=1}^{N} m_{i} r_{i}^{2} \alpha$$ The sum of all the terms \(m_{i} r_{i}^{2}\) represents the moment of inertia \(I\) of the rigid object: $$I = \sum_{i=1}^{N} m_{i} r_{i}^{2}$$ Now, we can substitute this expression for the moment of inertia into the sum of torques: $$\sum_{i=1}^{N} \tau_{i} = I \alpha$$ Thus, we have derived the rotational form of Newton's second law: $$\sum_{i=1}^{N} \tau_{i} = I \alpha$$ This equation shows that the net torque acting on a rigid object is equal to the product of its moment of inertia and angular acceleration.

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