ohgodamnitw0

2022-09-06

Do Newton's laws of motion apply on rigid bodies? If they apply on rigid bodies, would we consider forces acting in any direction or on any part of the body, and consider only the centre of mass when we talk about its momentum or the body being at rest or in uniform motion?

Because I used to regard Newton's laws as only applying to point masses, but I'm not sure if that's the case. For example, I read that the following statement can be justified using Newton's first law:

Consider a lever on a fulcrum with weights ${W}_{1}$ and ${W}_{2}$ on either side of the fulcrum, where the lever is in balance; the force exerted by the tip of the fulcrum on the lever is ${W}_{1}+{W}_{2}$

Because I used to regard Newton's laws as only applying to point masses, but I'm not sure if that's the case. For example, I read that the following statement can be justified using Newton's first law:

Consider a lever on a fulcrum with weights ${W}_{1}$ and ${W}_{2}$ on either side of the fulcrum, where the lever is in balance; the force exerted by the tip of the fulcrum on the lever is ${W}_{1}+{W}_{2}$

Conor Daniel

Beginner2022-09-07Added 11 answers

Absolutely Newton's laws apply for rigid bodies. There are extensions to F=ma attributed to Euler that describe the rotational equations of motion.

To be fully desciptive use point C to designate the center of mass and write

Momentum of body from the motion of the center of mass

$\begin{array}{}\text{(1)}& \overrightarrow{p}=m\phantom{\rule{thinmathspace}{0ex}}{\overrightarrow{v}}_{\mathrm{C}}\end{array}$

Newton's 2nd law as the time derivative of the above

$\begin{array}{}\text{(2)}& \overrightarrow{F}=m\phantom{\rule{thinmathspace}{0ex}}{\overrightarrow{a}}_{\mathrm{C}}\end{array}$

where $\overrightarrow{F}$ is the net force acting the body (applied anywhere, including external and reaction forces). Also ${a}_{\mathrm{C}}$ is the acceleration of the center of mass only.

Angular momentum about the center of mass is

$\begin{array}{}\text{(3)}& {L}_{\mathrm{C}}={\mathrm{I}}_{\mathrm{C}}\overrightarrow{\omega}\end{array}$

Where $\overrightarrow{\omega}$ is the rotationa velocity of the body (shared among all point on the body) and ${\mathrm{I}}_{C}$ is the mass moment of inertia (tensor) summed at the center of mass.

The motion about the center of mass is described by Euler law of rotation which is the time derivative of the above

$\begin{array}{}\text{(4)}& {M}_{\mathrm{C}}={\mathrm{I}}_{\mathrm{C}}\overrightarrow{\alpha}+\overrightarrow{\omega}\times {\mathrm{I}}_{\mathrm{C}}\overrightarrow{\omega}\end{array}$

Where $\overrightarrow{\alpha}$ is the rotational acceleration of the body and ${M}_{\mathrm{C}}$ the net torque about the center of mass.

All of the above can be easily derived when considering a rigid body as a collection of finite particles each moving with a translation of the center of mass and a rotation about the center of mass. Every "Introduction to Dynamics" book out there should have this somewhere in the first chapters.

To be fully desciptive use point C to designate the center of mass and write

Momentum of body from the motion of the center of mass

$\begin{array}{}\text{(1)}& \overrightarrow{p}=m\phantom{\rule{thinmathspace}{0ex}}{\overrightarrow{v}}_{\mathrm{C}}\end{array}$

Newton's 2nd law as the time derivative of the above

$\begin{array}{}\text{(2)}& \overrightarrow{F}=m\phantom{\rule{thinmathspace}{0ex}}{\overrightarrow{a}}_{\mathrm{C}}\end{array}$

where $\overrightarrow{F}$ is the net force acting the body (applied anywhere, including external and reaction forces). Also ${a}_{\mathrm{C}}$ is the acceleration of the center of mass only.

Angular momentum about the center of mass is

$\begin{array}{}\text{(3)}& {L}_{\mathrm{C}}={\mathrm{I}}_{\mathrm{C}}\overrightarrow{\omega}\end{array}$

Where $\overrightarrow{\omega}$ is the rotationa velocity of the body (shared among all point on the body) and ${\mathrm{I}}_{C}$ is the mass moment of inertia (tensor) summed at the center of mass.

The motion about the center of mass is described by Euler law of rotation which is the time derivative of the above

$\begin{array}{}\text{(4)}& {M}_{\mathrm{C}}={\mathrm{I}}_{\mathrm{C}}\overrightarrow{\alpha}+\overrightarrow{\omega}\times {\mathrm{I}}_{\mathrm{C}}\overrightarrow{\omega}\end{array}$

Where $\overrightarrow{\alpha}$ is the rotational acceleration of the body and ${M}_{\mathrm{C}}$ the net torque about the center of mass.

All of the above can be easily derived when considering a rigid body as a collection of finite particles each moving with a translation of the center of mass and a rotation about the center of mass. Every "Introduction to Dynamics" book out there should have this somewhere in the first chapters.

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force $\overrightarrow{F}$ The magnitude of the tension in the string between blocks B and C is T=3.00N. Each block has mass m=0.400kg.

What is the magnitude F of the force?

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???A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. On the rough section, the coefficient of friction is not constant but starts at .100 at P and increases linerly with distance past P, reaching a value of .600 at 12.5 m past point P. (a) Use the work energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid iff the friciton coefficient didn't increase, but instead had the constant value of .1?

?

The spring in the figure (a) is compressed by length delta x . It launches the block across a frictionless surface with speed v0. The two springs in the figure (b) are identical to the spring of the figure (a). They are compressed by the same length delta x and used to launch the same block. What is the block's speed now?A spring gun's spring has a constant force k =400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring.The ball is then launched out of the gun's barrel after the spring is released. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. Calculate the speed with which the ballleaves the barrel if you can ignore friction. Calculate the speed of the ball as it leavesthe barrel if a constant resisting force of 6.00 Nacts on the ball as it moves along the barrel. For the situation in part (b), at what position along the barrel does the ball have the greatest speed?

?A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 with the vertical. Air Resistance is negligible.

?

a) What is the speed of the rock when the string passesthrough the vertical position?

b) What is the tension in the string when it makes an angle of 45 with the vertical?

c) What is the tension in the string as it passes through the vertical?

Calculate the actual mechanical advantage of a lever.

The change in internal energy of a system that has absorbed 2 kcal of heat and does 500 J of work is

A) 6400J

B) 5400J

C) 7860J

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A) Path I

B) Path II

C) Same for both the paths

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