/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 A 55-kg runner runs around the e... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 87

A 55-kg runner runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 2.8 m/s. The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.20 rad/s relative to the earth. The radius of the turntable is 3.0 m, and its moment of inertia about the axis of rotation is 80 kg \(\cdot\) m\(^2\). Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can model the runner as a particle.)

Short Answer

Expert verified
The final angular velocity of the system is approximately 0.776 rad/s.

Step by step solution

01

Understanding Initial Conditions

Initially, the linear velocity of the runner relative to the earth is given as \(v_{r} = 2.8\,\text{m/s}\). The turntable has an initial angular velocity relative to the earth of \(\omega_{t} = -0.20\,\text{rad/s}\) (negative because it's in the opposite direction). We know the radius of the turntable \(r = 3.0\,\text{m}\), and the mass of the runner \(m = 55\,\text{kg}\). Also, the moment of inertia of the turntable is \(I = 80\,\text{kg}\,\cdot\,\text{m}^2\).
02

Calculate the Initial Angular Momentum of the Runner

The initial angular momentum of the runner \(L_r\) relative to the turntable can be calculated by \(L_r = m \cdot v_{r} \cdot r\) as the runner is modeled as a particle at a radius \(r\). Substituting the known values, we get \(L_r = 55 \cdot 2.8 \cdot 3.0 = 462\,\text{kg}\cdot\text{m}^2/\text{s}\).
03

Calculate the Initial Angular Momentum of the Turntable

The initial angular momentum of the turntable \(L_t\) is \(L_t = I \cdot \omega_t = 80 \cdot (-0.20) = -16\,\text{kg}\cdot\text{m}^2/\text{s}\). The negative sign indicates the direction is opposite to that of the runner.
04

Apply Conservation of Angular Momentum

According to the conservation of angular momentum, the total initial angular momentum must equal the total final angular momentum. Therefore, \(L_{initial} = L_{r} + L_{t} = 462 - 16 = 446\,\text{kg}\cdot\text{m}^2/\text{s}\). The final angular momentum \(L_{final}\) of the system is \( (I + m r^2) \cdot \omega_{final}\).
05

Solve for the Final Angular Velocity

Equating the initial and final angular momentum and solving for \(\omega_{final}\): \[ 446 = (80 + 55 \times 3.0^2) \cdot \omega_{final} \]\[ 446 = (80 + 495) \cdot \omega_{final} \]\[ 446 = 575 \cdot \omega_{final} \]\[ \omega_{final} = \frac{446}{575} \approx 0.776\,\text{rad/s}\]."}],

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Key Concepts

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

Rotational Dynamics
Rotational dynamics is the study of the motion of objects that rotate about an axis. This area of physics expands upon linear dynamics, which focuses on motion along a straight path. It plays a crucial role when dealing with rotating systems like the given turntable scenario. The key components include:
  • Angular velocity (\(\omega \)): This is the rate at which an object rotates. It's analogous to linear velocity in linear motion. For the turntable, the initial \(\omega\) was \(-0.20 \text{ rad/s} \), indicating rotation direction and speed.
  • Angular momentum (\(L\)): This measures the momentum of a rotating system and depends on the object's moment of inertia and angular velocity. Conservation of angular momentum is a pivotal principle in rotational dynamics, stating that in the absence of external torques, the total angular momentum remains constant.
By examining these components, we understand how the rotational dynamics of the runner and turntable interact. The problem illustrates how systems exchange angular momentum to adhere to the conservation law.
Moment of Inertia
Moment of inertia (\(I\)) is to rotational motion what mass is to linear motion. It quantifies how resistant an object is to rotational acceleration about a specific axis. In physics, it鈥檚 often referred to as rotational inertia.Here, the turntable has a moment of inertia of 80 \(\text{kg} \cdot \text{m}^2\). This value indicates how difficult it is to change its rotational speed. For the runner, modeled as a particle, the moment of inertia is computed with \(m r^2\), where \(m\) is the runner鈥檚 mass and \(r\) is the distance from the axis.
  • Particle approximation: Treating the runner as a particle simplifies calculations because we focus primarily on their distance from the axis (\(r\)) and mass (\(m\)).
  • Combined moment of inertia: When systems combine (like the runner stopping relative to the turntable), their moments of inertia aggregate as \(I + m r^2\).
Understanding moment of inertia is essential for determining how rotational movements adjust within combined systems.
Angular Velocity
Angular velocity is a fundamental parameter in rotational motion, describing how quickly an object spins around an axis. It's measured in radians per second (\(\text{rad/s}\)). This differs from linear velocity, which measures how fast something moves along a path.In the given problem, both the runner and the turntable possess their initial angular velocities with opposite directions:
  • Runner's angular velocity relative to Earth: Initially depicted by their linear velocity (\(2.8 \text{ m/s}\)). This must be converted to angular velocity by using the radius of the turntable.
  • Turntable's angular velocity: Given initially as \(-0.20 \text{ rad/s}\), it highlights the rotational speed and direction.
  • Final System Velocity: When the runner stops relative to the turntable, the system's final angular velocity is deduced by equating initial and final angular momentum. Resulting in a combined velocity that adapts to accommodate the runner's halt.
Angular velocity is vital in understanding how rotational speeds shift as systems interact and modify.

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Most popular questions from this chapter

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it have then?

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 N is applied to the rim of the wheel. The wheel has radius 0.120 m. Starting from rest, the wheel has an angular speed of 12.0 rev/s after 2.00 s. What is the moment of inertia of the wheel?

If the body's center of mass were not placed on the rotational axis of the turntable, how would the person's measured moment of inertia compare to the moment of inertia for rotation about the center of mass? (a) The measured moment of inertia would be too large; (b) the measured moment of inertia would be too small; (c) the two moments of inertia would be the same; (d) it depends on where the body's center of mass is placed relative to the center of the turntable.

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg \(\cdot\) m\(^2\) about a vertical axle through its center, and it turns with negligible friction. (a) A child applies an 18.0-N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest, what is its angular speed after this 15.0-s interval? (b) How much work did the child do on the merry-go-round? (c) What is the average power supplied by the child?
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