Newton’s Laws and Circular Motion

From Young and Freedman’s University Physics with Modern Physics Textbook, 14th edition

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A stone with mass 0.8 kg is attached to the end of 0.9 m string. The string will break if its tension exceeds 60 N. The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. (a) draw a free body diagram of the stone. (b) Find the maximum speed the stone can attain without the string breaking.

 

Analyze the problem:

A stone is tied into an end of a string and then swung horizontally to move in circle. As the stone move in circular motion then a radial acceleration is involved. The string can uphold a tension force up to 60 N so there will be a maximum radial acceleration that when exceeded the string will break.

What are the given information?

  1. The stone mass, m=0.8 kg
  2. The string length, L= 0.9 m
  3. Maximum tension, T=60 N

What are the needed information?

  1. Body-diagram for the stone.
  2. maximum speed of stone so the string is not breaking

SOLUTION:

NLCM1

a)NLCM2b)

NLCM3

Newton’s Laws Application

Source: Young and Freedman’s University Physics with Modern Physics textbook, 14th edition

Blocks A, B, and C are connected using rope of negligible mass (see figure below). Both A and B weight 23.8 N each, and the coefficient of kinetic friction between each block and the surface is 0.32. Block C descends with constant velocity.NLA1

a) Draw free body diagram for blocks A and B.

b) Find tension in rope connecting blocks A and B.

c) What is the weight of block C?

d) If the rope connecting A and B were cut, what would be the acceleration of block C?

Analyze the problem:

The problem has three blocks attached by an inelastic string (so they should have the same acceleration) and pulleys mass ignored. The surface where Blocks A and B are set on have friction and the kinetic friction coefficient was given (kinetic friction coefficient is involved when bodies are in motion which is different than static friction coefficient which used when bodies are not in motion).

What are the given information?

  1. Block A weight, FwA=23.8 N
  2. Block B weight, FwB=23.8 N
  3. Kinetic friction coefficient for all surfaces = 0.32
  4. Block C velocity is constant, so acceleration, a =0 m/s2

What are the needed information?

  1. Body-diagram draw for blocks A and B.
  2. Tension in rope attaching block A and block B.
  3. The weight of Block C
  4. When rope between block A and block B were cut , then find the acceleration of block C.

SOLUTION:

Note: The acceleration shown in figure below is for part (d).

NLA2

NLA3

NLA5

 

 

Two Blocks on Inclined plane

“Physics Principles with Applications”, Douglas C. Giancoli, 7th Edition, Global Edition

44. Two blocks are connected by a light string passing over a pulley of radius 0.15 m and moment of inertia I. The blocks move (toward the right) with an acceleration of 1.00 m/s2 along their frictionless inclines (See figure below). (a) Draw free-body diagrams for each of the two blocks and the pulley. (b) Determine FTA and FTB, the tensions in the two parts of the string. (c) Find the net torque acting on the pulley, and determine its moment of inertia, I.

Giancoli 8-44

Analyze the problem:

The problem has two blocks on inclined surfaces and they are attached by an inelastic string (so they should have the same acceleration) the pulley’s mass is not ignored (we know that because the problem mentioned that there is a moment of inertia).

What are the given information?

  1. Block A mass, mA=8 kg
  2. Block B mass, mB=10 kg
  3. Acceleration, a = 1 m/s2
  4. Inclination Angle 1, θ1=32º
  5. Inclination Angle 2, θ2=61º
  6. Radius of pulley, R= 0.15 m

What are the needed information?

  1. Body-diagram draw for the two blocks and the pulley.
  2. FTA and FTB tension forces
  3. The net torque on the pulley
  4. The moment of inertia of the pulley

SOLUTION:

Giancoli 8-44-1

Giancoli 8-44-2

Atwood Machine

Source:
“Physics Principles with Applications”, Douglas C. Giancoli, 7th Edition, Global Edition

This is a common physics problem and it mentioned in several textbooks.

45. An Atwood machine (shown in figure below) consists of two masses, mA=65 kg and mB=75 kg, connected by a massless inelastic cord that passes over a pulley free to rotate, The pulley is a solid cylinder of radius R=0.45 m and mass 6 kg. (a) Determine the acceleration of each mass. (b) What % error would be made if the moment of inertia of the pulley is ignored? [Hint: the tensions FTA and FTB aare not equal]

Analyze the problem:

It is an Atwood typical problem but here the pulley’s mass  is not ignored and the tensions forces are different for the same cord. Having the pulley’s mass ignored and the tension being the same for the single cord simplify the problem, so here we have to approach the problem in different way but using the same methods (Newton’s laws!).

What are the given information?

  1. Block A mass, mA = 65 Kg
  2. Block B mass, mB=75 Kg
  3. Pulley mass, M=6 kg
  4. Pulley Radius, R=0.45 m

What are needed information?

  1. Acceleration of both blocks.
  2. Percentage error of the two cases (the pulley’s moment of inertia is ignored and the pulley’s moment of inertia is not ignored).

 

SOLUTION:

  1. Finding the acceleration of the two blocks:

Giancoli 8-45

2. Finding the percentage error:

Giancoli 8-45-2

 

Newton’s Laws and Circular Motion

From Serway and Jewett Physics for Scientists and Engineers with modern physics, 9th edition

6.11 A 4.00-kg object is attached to a vertical rod by two strings, as in Figure bellow. The object rotates in a horizontal circle at constant speed 6.00 m/s. Find the tension in (a) the upper string and (b) the lower string.

Serway 6.11 problem.jpg

Analyze the problem:

The problems states that it wants tension force in string for the above setting. Newton’s laws should be used beside the dynamic of circular motion concepts.

What are given information:

  1. Strings lengths (L) = 2 m
  2. Height (h) = 3 m
  3. Mass of object (m) = 4 kg
  4. Speed of object (v) = 6 m/s

What the problem need:

  1. Tension of upper string
  2. Tension of lower string

 

SOLUTION:

serway-6-11-problem-sol-aFrom above figure we can see that we have isosceles triangle hence we can have right triangle by drawing a perpendicular line on the base of triangle to get sides lengths 2 m, and 3/2=1.5 m, so:

Serway 6.11 problem main sol.jpg

Then:

The tension in the upper string and  the lower string

Serway 6.11 problem main sol2.jpg

 

 

 

 

Simple Dynamics of Circular Motion Problem

From Serway and Jewett Physics for Scientists and Engineers with modern physics, 9th edition

Serway 6.8 problem.jpg

Let’s analyze the wording of the problem:

  1. A conical pendulum is a swinging pendulum that can rotate around an axis. The bob (the mass) will have circular path at all angles θ, but here we need to investigate when the angle is 5°.
  2. When a body moves is circular motion it must have at least a radial acceleration (centripetal acceleration).
  3. The mass is connected by a wire, the wire will have a tension (because it is used to suspend the bob). The tension should have x- and y-components.

What the information give in problem:

  1. The bob’s mass (m) = 80 kg
  2. Length of wire (L) = 10 m
  3. Angle of pendulum (θ) = 5°

What the problem need:

  1. The x- and y-components of tension force.
  2. The radial acceleration

 

SOLUTION:

  • The x- and y-components of tension force.
    • The x- and y-components of the tension force are:

serway-6-8-problem-sol-aa

  • The radial acceleration

Serway 6.8 problem -sol b.jpg

Newton’s Laws Problem

From Young and Freedman’s University Physics with Modern Physics Textbook, 14th edition

problem-4-28

Let’s analyze the wording of the problem:

  1. “…bullet…strikes a large tree…” and “…assume a constant retarding force…”, gives hints that Newton’s laws would be involved.
  2. “What force… the tree exert on the bullet?” means Newton’s 2nd and 3rd laws will be used.

What information given in this problem:

  1. Speed of bullet = 350 m/s
  2. Penetration depth = 0.13 m = distance traveled
  3. Bullet mass = 1.8 g
  4. 0.22-caliber rifle not an important information for solving the problem.

What the problem need:

  1. The time required for the bullet to stop inside the tree.
  2. The force the tree exert on the bullet.

SOLUTION:

problem-4-28-sol1