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A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a mass of \(150 \mathrm{~kg}\), a radius of \(2.0 \mathrm{~m}\), and \(a\) rotational inertia of \(300 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axis of rotation. A \(60 \mathrm{~kg}\) student walks slowly from the rim of the platform toward the center. If the angular speed of the system is \(1.5 \mathrm{rad} / \mathrm{s}\) when the student starts at the rim, what is the angular speed when she is \(0.50 \mathrm{~m}\) from the center?

Short Answer

Expert verified
The final angular speed is approximately 2.857 rad/s.

Step by step solution

01

Identify Given Information

The mass of the platform is 150 kg, radius is 2.0 m, and rotational inertia is 300 kg·m². The mass of the student is 60 kg. The initial angular speed is 1.5 rad/s. We need to find the angular speed when the student is 0.50 m from the center.
02

Apply Conservation of Angular Momentum

The system is isolated, so the angular momentum should be conserved. Let \(L_i\) be the initial angular momentum, and \(L_f\) be the final angular momentum when the student is 0.50 m from the center. Thus, \(L_i = L_f\).
03

Calculate Initial Angular Momentum

The initial angular momentum \(L_i\) is given by \(L_i = (I + m r^2) \omega_i\), where \(I = 300\, \mathrm{kg \cdot m^2}\), \(m = 60\, \mathrm{kg}\), \(r = 2.0\, \mathrm{m}\), and \(\omega_i = 1.5\, \mathrm{rad/s}\). Substitute these values to get \[ L_i = (300 + 60 \times 2^2) \times 1.5 = 900 \mathrm{\, kg \cdot m^2/s} \].
04

Calculate Final Angular Momentum

For the final state, the student is 0.50 m from the center, so the final angular momentum \(L_f = (I + m r_f^2) \omega_f\), where \(r_f = 0.50 \, \mathrm{m}\).
05

Solve for Final Angular Speed

Substitute \(L_i = L_f\) into the equation: \[ 900 = (300 + 60 \times 0.5^2) \times \omega_f \]. Solve for \(\omega_f\): \[ 900 = (300 + 60 \times 0.25) \times \omega_f \Rightarrow 900 = 315 \omega_f \Rightarrow \omega_f = \frac{900}{315} \approx 2.857 \mathrm{rad/s} \].

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Key Concepts

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

Rotational Inertia
Rotational inertia, also known as the moment of inertia, measures an object's resistance to changes in its rotational motion. In simpler terms, it's how "hard" it is to spin an object around a particular axis. The further mass is distributed from the axis, the higher the rotational inertia. For example, a heavy wheel takes more effort to spin if its mass is concentrated on the outer edge rather than close to the center. In equations, rotational inertia is represented by the symbol \(I\) and is calculated differently based on the object's shape and the axis of rotation. In the given problem, the rotational inertia of the platform is provided as \(300 \, \text{kg} \cdot \text{m}^2\). This value is crucial because it helps determine how the platform reacts as forces are applied or when objects (like the student) move on it. Understanding rotational inertia is key to solving problems where rotational motion and conservation principles are involved. It sets the foundation for calculating how different configurations—such as mass moving closer to or farther away from the center—affect angular speed.
Angular Speed
Angular speed is the rate at which an object rotates or revolves around an axis. It is analogous to linear speed but in a circular or rotational context. Represented with the Greek letter \(\omega\), angular speed is given in radians per second (rad/s). It tells you how fast an object is spinning.

In our scenario, the initial angular speed \(\omega_i\) is \(1.5 \, \text{rad/s}\) when the student starts at the rim of the platform. When the student walks towards the center, this affects the system's rotational inertia, leading to a change in angular speed. Since angular momentum is conserved in an isolated system (and our platform is frictionless and isolated), any change in the distribution of mass around the axis of rotation causes a change in angular speed to compensate, ensuring the product of rotational inertia and angular speed remains constant.

By understanding how changes in mass distribution affect angular speed, we can predict and calculate new conditions of motion for rotating systems, like the final angular speed \(\omega_f\) found in the exercise.
Physics Problem Solving
Physics problem solving often involves breaking down complex phenomena into fundamental principles and laws. For rotational motion problems like the one here, the conservation of angular momentum is a crucial concept. It's like peeling away layers to get to the core understanding.

Here's a typical approach:
  • Start by identifying all given information, such as masses, distances, and initial conditions.
  • Determine the physical laws applicable, such as conservation laws. In this case, angular momentum \(L\), given by \(L = I \cdot \omega\), must remain constant.
  • Translate the problem into mathematical equations using the identified laws.
  • Solve for the desired variable, which was the final angular speed \(\omega_f\) in the exercise, by manipulating the equations appropriately.

By systematically applying these steps, understanding becomes more manageable and solving becomes straightforward. Tackling physics problems by dissecting them into smaller, understandable parts helps in mastering various scenarios and gaining a deeper insight into how the physical world operates.

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Most popular questions from this chapter

A solid brass ball of mass \(0.280 \mathrm{~g}\) will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius \(R=14.0 \mathrm{~cm}\), and the ball has radius \(r \ll R\). (a) What is \(h\) if the ball is on the verge of leaving the track when it reaches the top of the loop? If the ball is released at height \(h=\) \(6.00 R\), what are the (b) magnitude and (c) direction of the horizontal force component acting on the ball at point \(Q\) ?

At the instant the displacement of a \(2.00 \mathrm{~kg}\) object relative to the origin is \(\vec{d}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}+(4.00 \mathrm{~m}) \hat{\mathrm{j}}-(3.00 \mathrm{~m}) \hat{\mathrm{k}}\), its veloc- ity is \(\vec{v}=-(6.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\) and it is subject to a force \(\vec{F}=(6.00 \mathrm{~N}) \hat{\mathrm{i}}-(8.00 \mathrm{~N}) \hat{\mathrm{j}}+(4.00 \mathrm{~N}) \hat{\mathrm{k}}\). Find (a) the accel- eration of the object, (b) the angular momentum of the object about the origin, (c) the torque about the origin acting on the object, and (d) the angle between the velocity of the object and the force acting on the object.

A small, solid, uniform ball is to be shot from point \(P\) so that it rolls smoothly along a horizontal path, up along a ramp, and onto a plateau. Then it leaves the plateau horizontally to land on a game board, at a horizontal distance \(d\) from the right edge of the plateau. The vertical heights are \(h_{1}=5.00 \mathrm{~cm}\) and \(h_{2}=1.60 \mathrm{~cm}\). With what speed must the ball be shot at point \(P\) for it to land at \(d=6.00 \mathrm{~cm} ?\)

A \(30 \mathrm{~kg}\) child stands on the edge of a stationary merry-go-round of radius \(2.0 \mathrm{~m}\). The rotational inertia of the merry-go-round about its rotation axis is \(150 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The child catches a ball of mass \(1.0 \mathrm{~kg}\) thrown by a friend. Just before the ball is caught, it has a horizontal velocity \(\vec{v}\) of magnitude \(12 \mathrm{~m} / \mathrm{s}\), at angle \(\phi=37^{\circ}\) with a line tangent to the outer edge of the merry-go-round, as shown. What is the angular speed of the merry-go-round just after the ball is caught?

An automobile traveling at \(80.0 \mathrm{~km} / \mathrm{h}\) has tires of \(75.0 \mathrm{~cm}\) diameter. (a) What is the angular speed of the tires about their axles? (b) If the car is brought to a stop uniformly in \(30.0\) complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the braking?

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