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Chapter 4: Problem 71

A circular platform is mounted on a vertical frictionless axle. Its radius is \(r=2 \mathrm{~m}\) and its moment of inertia is \(I=200 \mathrm{~kg} \mathrm{~m}^{2} .\) It is initially at rest. A \(70 \mathrm{~kg}\) man stands on the edge of the platform and begins to walk along the edge at speed \(v_{0}=10 \mathrm{~ms}^{-1}\) relative to the ground. The angular velocity of the platform is (1) \(1.2 \mathrm{rad} \mathrm{s}^{-1}\) (2) \(0.4 \mathrm{rad} \mathrm{s}^{-1}\) (3) \(2.0 \mathrm{rad} \mathrm{s}^{-1}\) (4) \(0.7 \mathrm{rad} \mathrm{s}^{-1}\)

Short Answer

Expert verified
The platform's angular velocity is calculated; however, results go beyond provided options, suggesting re-check needed.

Step by step solution

01

Understanding the Problem

We have a circular platform with a radius of \( r = 2 \) meters and a moment of inertia \( I = 200 \, \text{kg} \, \text{m}^2 \). A man with a mass \( m = 70 \, \text{kg} \) is walking around the edge of the platform at a speed of \( v_0 = 10 \, \text{m/s} \). We need to determine the angular velocity \( \omega \) of the platform.
02

Applying Conservation of Angular Momentum

The system is initially at rest, so initial angular momentum is zero. As the man walks, his linear momentum will create an angular momentum about the center. By conservation of angular momentum, initial total angular momentum (0) = man's angular momentum \( L_m \) + platform's angular momentum \( L_p \).
03

Calculating Man's Angular Momentum

The man's angular momentum \( L_m \) is given by \( m \times v_0 \times r \). With \( m = 70 \, \text{kg} \), \( v_0 = 10 \, \text{m/s} \) and \( r = 2 \, \text{m} \), we have: \( L_m = 70 \times 10 \times 2 = 1400 \, \text{kg} \, \text{m}^2/\text{s} \).
04

Calculating Platform's Angular Momentum

Platform's angular momentum \( L_p = I \times \omega \), where \( I = 200 \, \text{kg} \, \text{m}^2 \) and \( \omega \) is what we need to find. So \( L_p = 200 \cdot \omega \).
05

Setting Up Conservation Equation

According to conservation of angular momentum: \[ L_m + L_p = 0 \]Substituting the values we have: \[ 1400 - 200 \omega = 0 \]
06

Solving for Angular Velocity \(\omega\)

Solve the equation: \( 1400 - 200 \omega = 0 \) to find \( \omega \).Rearranging gives: \( 200 \omega = 1400 \), leading to \( \omega = \frac{1400}{200} = 7 \, \text{rad/s} \). Please remember to double-check this calculation. In the real setting, there seems to be an error in initial steps or book answers may differ due to conditions not mentioned.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
Imagine a figure skater spinning with their arms outstretched. As they pull their arms in, they spin faster. This happens because of the concept called **moment of inertia**, which is a measure of how difficult it is to change an object's rotation.
  • The moment of inertia depends on the mass of an object and how that mass is distributed relative to the axis of rotation.
  • For the circular platform in the exercise, it has a moment of inertia of 200 kg m虏. This means it requires a certain force to start or stop its rotation.
In practical terms, a larger moment of inertia means it's harder to change the rotational speed. That's why the platform's motion changes when the man starts to walk. His movement transfers angular momentum to the platform, causing it to rotate.
Circular Motion
When any object moves along a circular path, we're dealing with what we call **circular motion**.
  • This type of motion is characterized by a constantly changing direction, but the distance from the center remains the same.
  • Our man on the platform is an example where his motion in a circle affects the entire system.
Given that he walks around the edge of the platform, his linear motion is partly responsible for the platform鈥檚 rotation. Because he maintains a constant speed, he introduces a constant angular momentum to the system.
Angular Velocity
Now, let's talk about **angular velocity**, which is how fast something rotates or revolves, expressed in radians per second.
  • It's similar to linear speed but applies to rotation.
  • For the platform in the exercise, the man's movement creates angular velocity by changing the system's angular momentum.
In this context, by conservation of angular momentum, the platform gains an angular velocity opposite to the man's motion. The calculated angular velocity is obtained by equating the total angular momentum with the man's contribution and then solving for the platform's rotation.

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Most popular questions from this chapter

A uniform thin stick of mass \(M=24 \mathrm{~kg}\) and length \(L\) rotates on a frictionless horizontal plane, with its centre of mass stationary. A particle of mass \(m\) is placed on the plane at a distance \(x=L / 3\) from the centre of the stick. This stick hits the particle elastically. Find the value of \(m / M\) so that after the collision, there is no rotational motion of the stick.

A uniform solid sphere of radius \(r\) is rolling on a smooth horizontal surface with velocity \(V\) and angular velocity \(\omega=(V=\omega r) .\) The sphere collides with a sharp edge on the wall as shown in figure. The coefficient of friction between the sphere and the edge \(\mu=1 / 5\). Just after the collision the angular velocity of the sphere becomes equal to zero. The linear velocity of the sphere just after the collision is equal to (1) \(V\) (2) \(\frac{V}{5}\) (3) \(\frac{3 V}{5}\) (4) \(\frac{V}{6}\)

A uniform smooth rod (mass \(m\) and length \(l\) ) placed on a smooth horizontal floor is hit by a particle (mass \(m\) ) moving on the floor, at a distance \(l / 4\) from one end elastically \((e=1)\). The distance travelled by the centre of the rod after the collision when it has completed three revolutions will be (1) \(2 \pi l\) (2) cannot be determined (3) \(\pi l\) (4) none of these

Two cylinders having radii \(2 R\) and \(R\) and moment of inertia \(4 I\) and \(I\) about their central axes are supported by axles perpendicular to their planes. The large cylinder is jinitially rotating clockwise with angular velocity \(\omega_{0}\). The [finall cylinder is moved to the right until it touches the lage cylinder and is caused to rotate by the frictional force between the two. Eventually slipping ceases and the two ylinders rotate at constant rates in opposite directions. During this (1) angular momentum of system is conserved (2) kinetic energy is conserved (3) neither the angular momentum nor the kinetic energy is conserved (4) both the angular momentum and kinetic energy are conserved

A thick walled hollow sphere has outer radius \(R .\) It rolls down an inclined plane without slipping and its speed at the bottom is \(v .\) If the inclined plane is frictionless and the sphere slides down without rolling, its speed at the bottom is \(5 v / 4\). What is the radius of gyration of the sphere? (1) \(\frac{R}{\sqrt{2}}\) (2) \(\frac{R}{2}\) (3) \(\frac{3 R}{4}\) (4) \(\frac{\sqrt{3} R}{4}\)
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