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Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the theoretical treatment of the topic. The idea is to provide a verbose explanation of the solution, detailing the application of theory. Solution presented here, therefore, is treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Representative problems and their solutions

We discuss problems, whose analysis is suited to the technique of treating rolling motion as pure rotation. For this reason, questions are categorized in terms of the characterizing features pertaining to the questions :

  • Positions on the rolling body with a specified velocity
  • Velocity of a particle situated at a specified position
  • Distance covered by a particle in rolling
  • Kinetic energy of rolling

Position on the rolling body with specified velocity

Example 1

Problem : At an instant, the contact point of a rolling disk of radius “R” coincides with the origin of the coordinate system. If the disk rolls with constant angular velocity, “ω”, along a straight line, then find the position of a particle on the vertical diameter, whose velocity is 1/ 2 of the velocity with which the disk rolls.

Solution : Here, the particle on the vertical diameter moves with a velocity, which is 1/ 2 of that of the velocity of the center of mass. Now, velocity of center of mass is :

v C = ω R

Let the particle be at a distance “y” from the point of contact on the vertical diameter. Then, velocity of the particle is :

Velocity of a particle

Velocity of a particle on the vertical diameter

v = ω r = ω y

According to question,

v = v C 2

Putting values,

ω y = v C 2 = ω R 2

y = R 2

This result is expected from the nature of relation “ v = ω r ”. It is a linear relation for vertical distance. The velocity varies linearly with the vertical distance.

Example 2

Problem : At an instant, the contact point of a rolling disk of radius “R” coincides with the origin of the coordinate system. If the disk rolls with constant angular velocity, “ω”, along a straight line, then find the position of a particle on the rim of the disk, whose speed is same as the speed with which the disk rolls.

Solution : Here, the particle on the rim of the disk moves with the same velocity as that of the velocity of the center of mass. Now, velocity of center of mass is :

v C = ω R

Let the particle be at P(x,y) as shown in the figure. Then, velocity of the particle is :

Velocity of a particle

Velocity of a particle on the rim of the disk

v = 2 v C sin ( θ 2 ) = 2 ω R sin ( θ 2 )

According to question,

v = v C

Putting values,

2 v C sin ( θ 2 ) = 2 ω R sin ( θ 2 ) = ω R

sin ( θ 2 ) = 1 2 = sin 30 ° θ 2 = 30 ° θ = 60 °

x = - R sin 60 ° = - 3 R 2

y = R cos 60 ° = R 2

Since there are two such points on the rim on either side of the vertical line, the coordinates of the positions of the particles, having same speed as that of center of mass are :

- 3 R 2 , R 2 and 3 R 2 , R 2

Velocity of a particle situated at a specified position

Example 3

Questions & Answers

which colour has the shortest wavelength in the white light spectrum
Mustapha Reply
how do we add
Jennifer Reply
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
Abhyanshu Reply
x=5.8-3.22 x=2.58
sajjad
what is the definition of resolution of forces
Atinuke Reply
what is energy?
James Reply
Ability of doing work is called energy energy neither be create nor destryoed but change in one form to an other form
Abdul
motion
Mustapha
highlights of atomic physics
Benjamin
can anyone tell who founded equations of motion !?
Ztechy Reply
n=a+b/T² find the linear express
Donsmart Reply
أوك
عباس
Quiklyyy
Sultan Reply
Moment of inertia of a bar in terms of perpendicular axis theorem
Sultan Reply
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Yara Reply
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate
Rama Reply
A balloon is released from the ground which rises vertically up with acceleration 1.4m/sec^2.a ball is released from the balloon 20 second after the balloon has left the ground. The maximum height reached by the ball from the ground is
Lucky Reply
work done by frictional force formula
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Torque
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Why are we takingspherical surface area in case of solid sphere
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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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