<< Chapter < Page Chapter >> Page >

Relative motion of a girl along the rim of a disk

The girl begins to walk along the rim at a speed of 1 m/s relative to the ground in clockwise direction.

Solution : Here, mass is not distributed symmetrically about the axle. The weight of girl constitutes a torque perpendicular to the axis of rotation about the point at which axle is fixed on the ground. However, torque on axle by the ground prevents rotation due to the torque resulting from girl’s weight in vertical plane. Thus, there is no external torque on the system and, therefore, angular momentum of the system of "girl and disk-axle" is conserved.

Torque due to the weight of girl

The orque due to the weight of girl is balanced by restoring force of ground on the axle.

Applying law of conservation of angular momentum in vertical direction (axis of rotation), we have :

L i = L f

Both platform and the girl are initially stationary with respect to ground. As such, initial angular momentum of the system is zero. From conservation law, it follows that the final angular momentum of the system is also zero.

But final angular momentum has two constituents : (i) final angular momentum of the platform and (ii) final angular momentum of the girl. Sum of the two angular momentums should, therefore, be zero. Let “P” and “G” subscripts denote platform and girl respectively, then :

L f = L Pf + L Gf = 0

Now, MI of the circular platform about the axis of rotation is :

I P = M R 2 2 = 100 x 3 2 2 = 450 kg - m 2

The mass of the particles constituting the girl are at equal perpendicular distances from the axis of rotation. Its MI about the axis of rotation is :

I G = m R 2 = 60 x 3 2 = 540 kg - m 2

From the question, it is given that final speed of the girl around the rim of the platform is 1 m/s in clockwise direction. It means that girl has linear tangential velocity of 1 m/s with respect to ground. Her angular velocity, therefore, is :

ω Gf = - v R = - 1 3 rad / s

Negative sign indicates that the girl is rotating clockwise. Now, putting these values, we have :

L Pf + L Gf = I P ω Pf + I G ω Gf = 0

ω Pf = - - 1 x 540 3 x 450 = 0.4 rad / s

The positive value of final angular velocity of platform means that the platform rotates in anti-clockwise direction.

Note : Alternatively, we can find angular momentum of the girl, using the defining equation for angular momentum of a point (considering girl to be a particle like mass) as :

L Gf = - m v r = - 60 x 1 x 3 = - 180 kg - m s 2

Example 3

Problem : A thin rod is placed coaxially within a thin hollow tube, which lies on a smooth horizontal table. The rod, having same mass “M” and length “L” as that of tube, is free to move within the tube. The system of "tube and rod" is given an initial angular velocity, "ω", about a vertical axis at one of its end. Considering negligible friction between surfaces, find the angular velocity of the rod, when it just slips out of the tube.

Rotation of coaxial tube and rod

The rod is free to move within the tube.

Solution : The first question that we need to answer is “why does rod slip out of the tube?”. To understand the process, let us concentrate on the motion of the rod only. Each particle of the rod, say at the far end, tends to move straight tangentially (natural tendency). However, there is no radial force available that could bend its path toward the axis of rotation. Hence, the particle tends to keep its path in tangential direction. As such, the particles have the tendency to keep themselves away from the axis of rotation. Eventually, the rod slips away from the axis of rotation.

Questions & Answers

if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
Abhyanshu Reply
what is the definition of resolution of forces
Atinuke Reply
what is energy?
James Reply
can anyone tell who founded equations of motion !?
Ztechy Reply
n=a+b/T² find the linear express
Donsmart Reply
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
Sudeer Reply
Misthu Reply
Why are we takingspherical surface area in case of solid sphere
Saswat Reply
In all situatuons, what can I generalize?
Cart Reply
the body travels the distance of d=( 14+- 0.2)m in t=( 4.0 +- 0.3) s calculate it's velocity with error limit find Percentage error
Clinton Reply

Get the best Physics for k-12 course in your pocket!

Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Physics for k-12' conversation and receive update notifications?