Newton’s Laws and Circular Motion

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


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





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



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


The tension in the upper string and  the lower string

Serway 6.11 problem main sol2.jpg





Dynamics of Circular Motion Problem

Source: Past exam (Kuwait University) 2nd Midterm, Fall Semester 2015

A pilot of mass m=60 kg in a jet aircraft executes a loop-the-loop, as shown in the figure. The aircraft moves in a vertical circle of radius 800 m at a constant speed of 200 m/s. Determine the force (in N) exerted by the seat on the pilot at the top of the loop.


Let’s analyze the paragraph above:

  1. Aircraft doing a vertical circular loop “…aircraft executes a loop-the loop,…. moves in a vertical circle…”. Then the problem is a circular motion problem. Knowledge about radial acceleration should be present.
  2. “…The force exerted by the seat on the pilot…” means it is a normal force. Normal force commonly does not present alone.. we should think about other forces that are present as well.

What are the given information:

  1. Pilot mass (m) = 60 kg
  2. The circular loop radius (R) = 800 m
  3. The speed of the plane (v) = 200 m/s
  4. Tangential acceleration (at)=0 m/s2 (clue: the two words “constant speed”).

What the problem need:

  1. The normal force on the pilot at the top (shown in figure above)



The pilot diagram at top of the loop:


The normal force (n) directed from seat to pilot, hence when the aircraft at top of loop and pilot is up side down the normal force is directed downward.

The pilot weight (Fw) directed from pilot toward the ground, hence it is directed downward.

From the below image, with a constant speed, the total acceleration rises from the circular motion is the radial acceleration (ar).






Circular motion with constant acceleration problem

This problem from Serway and Jewett’s Physics for scientists and engineers with modern physics textbook (9th edition)


Let’s analyze the wording of this problem:

  1. A ball swings in a vertical circle… then it is a circular motion problem.
  2. A given acceleration at specific position in space which is 36.9° past the lowest point on way up for the ball..then the given acceleration does not seem not to be a radial acceleration (see the diagram below in solution).

The information given in this problem:

  1. Radius of circular motion is 1.5 m (R=1.5m).
  2. The acceleration of the ball is a=-22.5iˆ+20.2jˆ m/s² at 36.9° away from the vertical axis (y-axis).

What the problem need:

  1. Sketch the vector diagram showing the components of the acceleration.
  2. Find the magnitude of radial acceleration.
  3. Find the speed and velocity of the ball.



  1. Sketch the vector diagram showing the components of the acceleration.

The diagram of acceleration vector


2. Find the magnitude of radial acceleration.

The magnitude of radial acceleration could be found by using the horizontal and vertical components of the given acceleration:


3. Find the speed and velocity of the ball.

The speed could be found using the radial acceleration and the radius of the circular motion:

problem 42 circular motion (c).jpg