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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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Most popular questions from this chapter

A diver can change his rotational inertia by drawing his arms and legs close to his body in the tuck position. After he leaves the diving board (with some unknown angular velocity), he pulls himself into a ball as closely as possible and makes 2.00 complete rotations in \(1.33 \mathrm{s}\). If his rotational inertia decreases by a factor of 3.00 when he goes from the straight to the tuck position, what was his angular velocity when he left the diving board?
A ceiling fan has four blades, each with a mass of \(0.35 \mathrm{kg}\) and a length of \(60 \mathrm{cm} .\) Model each blade as a rod connected to the fan axle at one end. When the fan is turned on, it takes 4.35 s for the fan to reach its final angular speed of 1.8 rev/s. What torque was applied to the fan by the motor? Ignore torque due to the air.
A solid sphere of mass 0.600 kg rolls without slipping along a horizontal surface with a transnational speed of \(5.00 \mathrm{m} / \mathrm{s} .\) It comes to an incline that makes an angle of \(30^{\circ}\) with the horizontal surface. Ignoring energy losses due to friction, to what vertical height above the horizontal surface does the sphere rise on the incline?

A man is doing push-ups. He has a mass of \(68 \mathrm{kg}\) and his center of gravity is located at a horizontal distance of \(0.70 \mathrm{m}\) from his palms and \(1.00 \mathrm{m}\) from his feet. Find the forces exerted by the floor on his palms and feet.
(a) Assume the Earth is a uniform solid sphere. Find the kinetic energy of the Earth due to its rotation about its axis. (b) Suppose we could somehow extract \(1.0 \%\) of the Earth's rotational kinetic energy to use for other purposes. By how much would that change the length of the day? (c) For how many years would \(1.0 \%\) of the Earth's rotational kinetic energy supply the world's energy usage (assume a constant \(1.0 \times 10^{21} \mathrm{J}\) per year)?
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