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A merry-go-round with a moment of inertia equal to \(1260 kg\cdot m^2\) and a radius of 2.5 m rotates with negligible friction at 1.70 rad/s. A child initially standing still next to the merry-go-round jumps onto the edge of the platform straight toward the axis of rotation, causing the platform to slow to 1.35 rad/s. What is her mass?

Short Answer

Expert verified
The child's mass is approximately 52.26 kg.

Step by step solution

01

Identify Initial Conditions

The merry-go-round has a moment of inertia of \(I = 1260 \, kg \cdot m^2\) and is rotating at an angular velocity \(\omega_i = 1.70 \, rad/s\). The radius of the merry-go-round is \(r = 2.5 \, m\).
02

Analyze Final Conditions

When the child jumps onto the merry-go-round, the system slows to a new angular velocity \(\omega_f = 1.35 \, rad/s\). We need to find the mass of the child, denoted as \(m\).
03

Apply Conservation of Angular Momentum

Angular momentum is conserved because there is no external torque. The initial angular momentum \(L_i = I \cdot \omega_i\), and the final angular momentum \(L_f = (I + m \cdot r^2) \cdot \omega_f\). Set these equal: \[ I \cdot \omega_i = (I + m \cdot r^2) \cdot \omega_f \]
04

Solve Equation for Child's Mass

Substitute known values and solve for \(m\): \[ 1260 \, kg \cdot m^2 \times 1.70 \, rad/s = (1260 \, kg \cdot m^2 + m \times (2.5 \, m)^2) \times 1.35 \, rad/s \]Rearrange to find \(m\): \[ 1260 \times 1.70 = 1.35 \times (1260 + m \times 6.25) \]\[ 2142 = 1.35 \times (1260 + 6.25m) \]\[ 2142 = 1701 + 8.4375m \]Subtract 1701 from both sides:\[ 441 = 8.4375m \]Finally, solve for \(m\) by dividing both sides by 8.4375:\[ m \approx 52.26 \, kg \]
05

Conclusion

The mass of the child is approximately \(52.26 \, kg\).

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Key Concepts

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

Moment of Inertia
The term "moment of inertia" is pivotal when dealing with rotational motion. It essentially measures an object's resistance to changes in its rotation and depends significantly on the mass distribution relative to the rotational axis. If you think of it like rotational mass, you'll get the basic idea.
In mathematical terms, for a solid object, the moment of inertia, often denoted by \( I \), is calculated using the integral forms of mass elements \((dm)\) at distances \( r \) from the axis of rotation. For series of discrete masses, it is calculated as \( I = \sum m_i r_i^2 \).
For a merry-go-round, the moment of inertia defines how difficult it is to twist or turn the ride. The larger the value, the more the force required to change its angular velocity. In this exercise, the moment of inertia is given as \( 1260 \, kg \cdot m^2 \), providing a base to comprehend how mass and rotation interact.
Understanding moment of inertia helps in various applications, from designing gears and engines to predicting the motion of celestial bodies.
Conservation of Angular Momentum
In physics, the principle of "conservation of angular momentum" is a fundamental concept stating that if no external torque acts on a system, the total angular momentum remains constant over time. This principle is analogous to the conservation of linear momentum but applies to rotational motion.
Angular momentum is represented as the product of the moment of inertia and angular velocity: \( L = I \cdot \omega \). In this exercise, before the child jumps onto the merry-go-round, the system has an initial angular momentum, \( L_i = I \cdot \omega_i \). After the child climbs aboard, the system's final angular momentum is \( L_f = (I + m \cdot r^2) \cdot \omega_f \).
Setting \( L_i \) equal to \( L_f \) because there is no external torque influencing the merry-go-round, allows us to calculate the child's mass. This relationship reflects how the two components, the moment of inertia and angular velocity, must adjust to maintain a constant state of angular motion.
Understanding this principle not only aids in solving theoretical problems but also provides insight into real-world phenomena like why a figure skater spins faster when pulling in their arms.
Angular Velocity
Angular velocity describes how fast an object rotates or revolves relative to another point, typically the center of rotation. It is a vector quantity, meaning it has both magnitude and direction.
Measured in radians per second (rad/s), angular velocity \( \omega \) indicates the rate of change of an angle. For the merry-go-round example, its initial angular velocity is \( 1.70 \, rad/s \), which decreases to \( 1.35 \, rad/s \) after the child adds their mass to the system.
The reduction in angular velocity illustrates the inverse relationship between an increased moment of inertia and angular speed when no external torque is present. Understanding this relationship is vital for controlling movements in devices and systems that rotate, from simple playground rides to complex machinery.
  • **Initial Angular Velocity (\( \omega_i \))**: Determines how quickly the ride rotates before the child jumps.
  • **Final Angular Velocity (\( \omega_f \))**: Slows due to the addition of the child’s mass, demonstrating conservation principles in action.
  • **Practical Implications**: Helps in designing and managing rotating systems, ensuring stability and efficiency.
Grasping angular velocity is key to mastering many physics and engineering topics.

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Most popular questions from this chapter

(I) \((a)\) What is the angular momentum of a 2.8-kg uniform cylindrical grinding wheel of radius 28 cm when rotating at 1300 rpm? \((b)\) How much torque is required to stop it in 6.0 s?

(II) The bolts on the cylinder head of an engine require tightening to a torque of 95 m\(\cdot\)N. If a wrench is 28 cm long, what force perpendicular to the wrench must the mechanic exert at its end? If the six-sided bolt head is 15 mm across (Fig. 8-44), estimate the force applied near each of the six points by a wrench.

(II) A rotating uniform cylindrical platform of mass 220 kg and radius 5.5 m slows down from 3.8 rev/s to rest in 16 s when the driving motor is disconnected. Estimate the power output of the motor (hp) required to maintain a steady speed of 3.8 rev/s.

(II) In traveling to the Moon, astronauts aboard the \(Apollo\) spacecraft put the spacecraft into a slow rotation to distribute the Sun's energy evenly (so one side would not become too hot). At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis. Determine \((a)\) the angular acceleration, and \((b)\) the radial and tangential components of the linear acceleration of a point on the skin of the ship 6.0 min after it started this acceleration.

Suppose David puts a 0.60-kg rock into a sling of length 1.5 m and begins whirling the rock in a nearly horizontal circle, accelerating it from rest to a rate of 75 rpm after 5.0 s. What is the torque required to achieve this feat, and where does the torque come from?

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