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Chapter 10: Problem 24

A girl of mass \(50.6 \mathrm{~kg}\) stands on the edge of a frictionless merry- go-round of mass \(827 \mathrm{~kg}\) and radius \(3.72 \mathrm{~m}\) that is not moving. She throws a \(1.13-\mathrm{kg}\) rock in a horizontal direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is \(7.82 \mathrm{~m} / \mathrm{s}\). Calculate \((a)\) the angular speed of the merry-go-round and \((b)\) the linear speed of the girl after the rock is thrown. Assume that the merry-go-round is a uniform disk.

Short Answer

Expert verified
The angular speed of the merry-go-round and the linear speed of the girl after the rock is thrown can be calculated using the principles of conservation of angular momentum and the formula for the angular momentum of a solid disc.

Step by step solution

01

Find initial total angular momentum

Before the rock is thrown, the system (merry-go-round, the girl, and the rock) is at rest, so the initial total angular momentum (L) is 0.
02

Find final angular momentum of the rock

After the rock is thrown, we need to calculate the angular momentum of the thrown rock. The angular momentum (L) of an object moving in a circle of radius r is given by \(L = mvr\), where m is the mass, v is the linear speed of the object. In this case: \(L = (1.13 kg) * (7.82 m/s) * (3.72 m)\).
03

Find final angular momentum of the girl and the merry-go-round

Due to conservation of angular momentum, the final total angular momentum of the system remains 0. Hence, the merry-go-round and the girl together have an angular momentum equal in magnitude and opposite in direction to the angular momentum of the rock. Their final angular momentum is \(-(1.13 kg) * (7.82 m/s) * (3.72 m)\).
04

Calculate the angular speed of the merry-go-round

The merry-go-round and the girl together can be treated as a solid disk. The angular momentum of a solid disk is \(L = Iω\), where I is the moment of inertia and ω is the angular speed. For a solid disk, \(I = 0.5*M*R^2\), where M is the total mass (mass of the merry-go-round plus mass of the girl) and R is the radius. So we can setup the equation: \(-(1.13 kg) * (7.82 m/s) * (3.72 m) = 0.5*(827 kg + 50.6 kg)*(3.72 m)^2*ω\). Solving for ω will give us the angular speed of the merry-go-round.
05

Calculate the linear speed of the girl

The linear speed (v) of an object moving in a circle equals the product of the radius (r) and the angular speed (ω), given by \(v = r * ω\). Using the value of ω calculated at previous step will give us the linear speed of the girl.

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Key Concepts

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

Angular Momentum
Angular momentum is a physical quantity that represents the rotational inertia of an object in motion around a fixed point or axis. Mathematically, it is the product of an object's mass, its linear velocity, and the perpendicular distance from the axis of rotation, which in the case of circular motion corresponds to the radius of the circle. It's expressed with the formula:
\[\begin{equation} L = mvr \end{equation}\] where
    \t
  • \t\tm is the mass of the object, \t
    • \t
  • \t\tv is the linear speed of the object, \t
    • \t
  • \t\tr is the radius of the path. \t
In the provided exercise, the concept of conservation of angular momentum is critical, which indicates that in the absence of external torques, the total angular momentum of a closed system remains constant. Thus, even though the girl on the merry-go-round changes the system's dynamics by throwing the rock, the overall angular momentum before and after the throw must remain zero.
Moment of Inertia
The moment of inertia is analogous to mass in linear motion, but instead, it measures an object's resistance to changes in its rotational motion. For a given object, it depends on both its mass distribution and the axis about which it rotates. A standard formula for the moment of inertia of a solid disk about its central axis is:
\[\begin{equation} I = 0.5 \times M \times R^2 \end{equation}\] where
    \t
  • \t\tM is the mass of the disk, \t
    • \t
  • \t\tR is the radius. \t
In our exercise, the moment of inertia is crucial for understanding how the merry-go-round's rotational speed responds after the girl throws the rock. It helps us calculate the new angular speed by relating the moment of inertia and the angular momentum of the system.
Angular Speed
Angular speed describes how fast an object rotates or revolves relative to another point, usually the center of a circle. It is the rate at which an angle is swept and is measured in radians per second. The formula connecting angular momentum (\(L\)), moment of inertia (\(I\)), and angular speed (\(\omega\)) is:\[\begin{equation} L = I\omega \end{equation}\]In the case of the girl on the merry-go-round, once we've understood the changes in angular momentum caused by throwing the rock, we can find the merry-go-round's angular speed. The final angular speed is a direct result of the conservation of angular momentum and is inversely proportional to the moment of inertia.
Linear Speed
While angular speed refers to how quickly an object spins around an axis, linear speed is about how quickly it moves along a straight path or, in the case of circular motion, the path's circumference. The formula to convert angular speed (\(\omega\)) to linear speed (\(v\)) in uniform circular motion is:\[\begin{equation} v = r \times \omega \end{equation}\]Here, the radius (\(r\)) is a bridge between the rotational motion and the motion along the circle's edge. After calculating the angular speed of the merry-go-round system, multiplying it by the radius gives us the linear speed of the girl as she rotates with the merry-go-round.
Uniform Circular Motion
Uniform circular motion describes an object moving in a perfect circle at a constant speed. Even though the speed is constant, because the direction of the motion is continually changing, there is a continuous acceleration towards the center of the circle, known as centripetal acceleration. This situation is distinct because, even though speed is constant, velocity is not – because velocity is a vector quantity that considers both speed and direction.

In our scenario, if the merry-go-round were to rotate at a constant angular speed without the influence of external forces or torques, the girl's motion on it would exemplify uniform circular motion. This allows us to straightforwardly apply the uniform circular motion equations to determine her linear speed and explore the underlying physical principles at play during her rotation on the merry-go-round.

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Most popular questions from this chapter

A particle of mass \(13.7 \mathrm{~g}\) is moving with a constant velocity of magnitude \(380 \mathrm{~m} / \mathrm{s}\). The particle, moving in a straight line, passes within \(12 \mathrm{~cm}\) of the origin. Calculate the angular momentum of the particle about the origin.

Let \(\overrightarrow{\mathbf{r}}_{\mathrm{cm}}\) be the position vector of the center of mass \(C\) of a system of particles with respect to the origin \(O\) of an inertial reference frame, and let \(\overrightarrow{\mathbf{r}}_{i}^{\prime}\) be the position vector of the \(i\) th particle, of mass \(m_{i}\), with respect to the center of mass \(C\). Hence \(\overrightarrow{\mathbf{r}}_{i}=\overrightarrow{\mathbf{r}}_{\mathrm{cm}}+\overrightarrow{\mathbf{r}}_{i}^{\prime}\) (see Fig. \(\left.10-22\right)\). Now define the total angular momentum of the system of particles relative to the center of mass \(C\) to be \(\overrightarrow{\mathbf{L}}^{\prime}=\Sigma\left(\overrightarrow{\mathbf{r}}_{i}^{\prime} \times \overrightarrow{\mathbf{p}}_{i}^{\prime}\right), \quad\) where \(\overrightarrow{\mathbf{p}}_{i}^{\prime}=m_{i} d \overrightarrow{\mathbf{r}}_{i}^{\prime} / d t\) (a) Show that \(\overrightarrow{\mathbf{p}}_{i}^{\prime}=m_{i} d \overrightarrow{\mathbf{r}}_{i} / d t-\) \(m_{i} d \overrightarrow{\mathbf{r}}_{\mathrm{cm}} / d t=\overrightarrow{\mathbf{p}}_{i}-m_{i} \overrightarrow{\mathbf{v}}_{\mathrm{cm}}\) (b) Show next that \(d \overrightarrow{\mathbf{L}}^{\prime} / d t=\) \(\Sigma\left(\overrightarrow{\mathbf{r}}_{i}^{\prime} \times d \overrightarrow{\mathbf{p}}_{i}^{\prime} / d t\right) .(c)\) Combine the results of \((a)\) and \((b)\) and, using the definition of center of mass and Newton's third law, show that \(\vec{\tau}_{\mathrm{ext}}^{\prime}=d \overrightarrow{\mathbf{L}}^{\prime} / d t\), where \(\vec{\tau}_{\mathrm{ext}}^{\prime}\) is the sum of all the \(\mathrm{ex}\) - ternal torques acting on the system about its center of mass.

A wheel of radius \(24.7 \mathrm{~cm}\), moving initially at \(43.3 \mathrm{~m} / \mathrm{s}\), rolls to a stop in \(225 \mathrm{~m} .\) Calculate \((a)\) its linear acceleration and \((b)\) its angular acceleration. ( \(c\) ) The wheel's rotational inertia is \(0.155 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) Calculate the torque exerted by rolling friction on the wheel.

Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.

Astronomical observations show that from 1870 to 1900 the length of the day increased by about \(6.0 \times 10^{-3}\) s. (a) What corresponding fractional change in the Earth's angular velocity resulted? \((b)\) Suppose that the cause of this change was a shift of molten material in the Earth's core. What resulting fractional change in the Earth's rotational inertia could account for the answer to part \((a) ?\)
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