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A merry-go-round with a moment of inertia equal to \(1260 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and a radius of \(2.5 \mathrm{~m}\) rotates with negligible friction at \(1.70 \mathrm{rad} / \mathrm{s}\). 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 \(1.25 \mathrm{rad} / \mathrm{s}\). What is her mass?

Short Answer

Expert verified
The child's mass is approximately 57.6 kg.

Step by step solution

01

Understand the Conservation of Angular Momentum

Angular momentum must be conserved when there is no external torque acting on the system. When the child jumps onto the merry-go-round, the total angular momentum of the system (merry-go-round + child) before and after must be equal. The equation can be expressed as \( L_{i} = L_{f} \), where \( L_{i} \) is the initial angular momentum and \( L_{f} \) is the final angular momentum.
02

Calculate Initial Angular Momentum

The initial angular momentum \( L_{i} \) of the system can be calculated using the formula: \( L_{i} = I_{m} \cdot \omega_{i} \), where \( I_{m} \) is the moment of inertia of the merry-go-round (\( 1260 \mathrm{~kg} \cdot \mathrm{m}^{2} \)) and \( \omega_{i} \) is the initial angular velocity (\( 1.70 \mathrm{rad} / \mathrm{s} \)). Therefore, \( L_{i} = 1260 \cdot 1.70 \).
03

Calculate Final Angular Momentum

After the child jumps on, the system's final angular momentum \( L_{f} \) can be expressed as the sum of the angular momentum of the merry-go-round and the child: \( L_{f} = I_{m} \cdot \omega_{f} + m \cdot r^2 \cdot \omega_{f} \). Here, \( \omega_{f} \) is the final angular velocity (\( 1.25 \mathrm{rad} / \mathrm{s} \)), \( m \) is the mass of the child, and \( r \) is the radius of the merry-go-round (\( 2.5 \mathrm{~m} \)). Simplifying, we have \( L_{f} = (I_{m} + m \cdot r^2) \cdot \omega_{f} \).
04

Equate and Solve for Child's Mass

Setting the initial and final angular momentum equal, we have: \( 1260 \cdot 1.70 = (1260 + m \cdot 2.5^2) \cdot 1.25 \). Simplifying the equation: \( 2142 = (1260 + 6.25m) \cdot 1.25 \). Solving for \( m \) gives \( m = \frac{2142 / 1.25 - 1260}{6.25} \). Calculating gives \( m \approx 57.6\). So, the child's mass is approximately \( 57.6 \) kg.

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Key Concepts

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

Conservation of Angular Momentum
In physics, the principle of the conservation of angular momentum is a pivotal concept. It states that if no external torque acts on a system, the total angular momentum remains constant. Angular momentum, a vector quantity, depends on the moment of inertia and angular velocity. It is defined as the product of the rotational inertia and angular velocity of a system. For a closed system, the sum of the angular momentum before any interaction, such as a child jumping onto a merry-go-round, must equal the sum afterward.
- **Initial Angular Momentum**: The angular momentum before an event like a jump.- **Final Angular Momentum**: The angular momentum after an event.In the given problem, when the child jumps onto the merry-go-round, this principle ensures that the system's initial angular momentum equals the final angular momentum. This is expressed mathematically as \( L_{i} = L_{f} \), which helped us solve for the mass of the child in the problem.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotation. Sometimes thought of as rotational mass, it plays a similar role in rotational dynamics as mass does in linear dynamics. The larger the moment of inertia, the more torque is required to change the angular velocity.
The moment of inertia depends on:
  • The mass of the object
  • The distribution of mass relative to the axis of rotation
In simpler terms, further mass from the axis increases the moment of inertia. For a solid disk like a merry-go-round, the moment of inertia is given by:\[ I = \frac{1}{2} m r^2 \]In this problem, the merry-go-round's moment of inertia is provided and crucial for calculating the system's initial angular momentum. Understanding this property helps us predict how objects will behave when rotated.
Angular Velocity
Angular velocity is the rate of change of angular position of a rotating object. It indicates how fast an object spins around an axis, measured in radians per second. In a practical context, it tells us the speed of rotation.
When a child jumps onto a merry-go-round, she affects its angular velocity. Before the jump, the merry-go-round spins at 1.70 rad/s, and 1.25 rad/s after. These values are crucial in setting up the conservation equation for angular momentum. Angular velocity is linked with:
  • Total angular momentum
  • Moment of inertia
For instance, if the moment of inertia increases as the child boards, and given that angular momentum is conserved, the angular velocity must decrease. Hence, understanding the relationship between these elements helps predict and explain the system's behavior during rotational collisions.
Physics Problem-Solving
Approach physics problems systematically to find clear solutions. Here's a simple strategy applied to the merry-go-round problem:
1. **Identify Known and Unknown Variables**: List the given data, like the initial and final angular velocities, radius, and moment of inertia. Determine what you're solving for (the child's mass).
2. **Apply Relevant Physics Concepts**: Use principles like the conservation of angular momentum to relate the known and unknowns. For this problem, set initial equal to final angular momentum.
3. **Formulate Equations**: Write equations based on relevant principles. In this case, use \( L_i = L_f \) and substitute known values for calculations.
4. **Solve**: Arrive at the unknown by performing algebraic manipulations. This involves simple calculations once the equation is set.
5. **Verify Solutions**: Check answers by verifying the logical consistency with physical principles. Ensure that calculations align with the law of conservation.Using these steps ensures clarity and effectiveness in approaching physics problems, enabling a logical journey from problem to solution.

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Most popular questions from this chapter

(II) A person stands on a platform, initially at rest, that can rotate frecly without friction. The moment of inertia of the person plus the platform is \(I_{\mathrm{P}}\) . The person holds a spinning bicycle whecl with its axis horizontal. The whecl has moment of inertia $$I_{\mathrm{w}} \text { and angular velocity } \omega_{\mathrm{W}} $$ What will be the angular velocity $$ \omega_{\mathbf{Y}} $$ of the platform if bthe person moves the axis of the wheel so that it points (a) vertically upward, (b) at a \(60^{\circ}\) angle to the vertical, \((c)\) vertically downward? (d) What will \(\omega_{\mathrm{P}}\) be if the person reaches up and stops the whecl in part \((a) ?\)

The position of a particle with mass \(m\) traveling on a helical path (see Fig. \(11-45\) ) is given by $$ \overrightarrow{\mathbf{r}}=R \cos \left(\frac{2 \pi z}{d}\right) \hat{\mathbf{i}}+R \sin \left(\frac{2 \pi z}{d}\right) \hat{\mathbf{j}}+z \hat{\mathbf{k}} $$ where \(R\) and \(d\) are the radius and pitch of the helix, respectively, and \(z\) has time dependence \(z=v_{z} t\) where \(v_{z}\) is the (constant) component of velocity in the \(z\) direction. Determine the time-dependent angular momentum \(\overrightarrow{\mathbf{L}}\) of the particle about the origin.

(II) (a) Show that the cross product of two vectors, $$ \begin{array}{c} \overrightarrow{\mathbf{A}}=A_{x} \hat{\mathbf{i}}+A_{y} \hat{\mathbf{j}}+A_{z} \hat{\mathbf{k}}, \text { and } \overrightarrow{\mathbf{B}}=B_{x} \hat{\mathbf{i}}+B_{y} \hat{\mathbf{j}}+B_{z} \hat{\mathbf{k}} \text { is } \\ \overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}=\left(A_{y} B_{z}-A_{z} B_{y}\right) \hat{\mathbf{i}}+\left(A_{z} B_{x}-A_{x} B_{z}\right) \hat{\mathbf{j}} \\ +\left(A_{x} B_{y}-A_{y} B_{x}\right) \hat{\mathbf{k}} \end{array} $$ (b) Then show that the cross product can be written $$ \overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \end{array}\right| $$ where we use the rules for evaluating a determinant. (Note, however, that this is not really a determinant, but a memory aid.)

A person of mass \(75 \mathrm{~kg}\) stands at the center of a rotating merry- go-round platform of radius \(3.0 \mathrm{~m}\) and moment of inertia \(920 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) The platform rotates without friction with angular velocity \(0.95 \mathrm{rad} / \mathrm{s}\). The person walks radially to the edge of the platform. (a) Calculate the angular velocity when the person reaches the edge. ( \(b\) ) Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.

$$ \vec{\mathbf{A}} \text { and } \vec{\mathbf{B}} $$ $$ \vec{\mathbf{A}} \times \vec{\mathbf{B}} $$ $$\vec{\mathbf{B}} \text { points south. }(b) \overline{\mathbf{A}} \text { points cast, } \vec{\mathbf{B}} $$ straight down. (c) \(\overline{\mathbf{A}}\) points straight up, \(\vec{\mathbf{B}}\) points north. (d) \(\vec{\mathbf{A}}\) points straight up, \(\vec{\mathbf{B}}\) points straight down.

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