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A horizontal vinyl record of mass \(0.10 \mathrm{~kg}\) and radius \(0.10 \mathrm{~m}\) rotates freely about a vertical axis through its center with an angular speed of \(4.7 \mathrm{rad} / \mathrm{s}\). The rotational inertia of the record about its axis of rotation is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2} .\) A wad of wet putty of mass \(0.020 \mathrm{~kg}\) drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately after the putty sticks to it?

Short Answer

Expert verified
The final angular speed is approximately 3.36 rad/s.

Step by step solution

01

Understand the Problem

The problem involves conservation of angular momentum. Initially, the system consists of only the vinyl record rotating freely, and then a wad of putty drops and sticks to the edge, altering the system's angular momentum. We need to find the new angular speed after the putty sticks.
02

Identify the Initial Angular Momentum

The initial angular momentum of the system is due to the rotating record. This is given by the formula \( L_i = I_i \omega_i \), where \( I_i = 5.0 \times 10^{-4} \; \mathrm{kg \cdot m^2}\) is the moment of inertia of the record, and \( \omega_i = 4.7 \; \mathrm{rad/s} \) is the initial angular speed. Calculate \(L_i = (5.0 \times 10^{-4}) \times 4.7 = 2.35 \times 10^{-3} \; \mathrm{kg \cdot m^2/s} \).
03

Determine the Final Moment of Inertia

When the putty sticks to the edge of the record, it changes the system's total moment of inertia. The putty contributes an additional moment of inertia given by \( I_{\text{putty}} = m r^2 \), where \( m = 0.020 \; \mathrm{kg} \) is the mass of the putty and \( r = 0.10 \; \mathrm{m} \) is the radius of the record. Calculate \(I_{\text{putty}} = 0.020 \times (0.10)^2 = 2.0 \times 10^{-4} \; \mathrm{kg \cdot m^2} \). The total final moment of inertia \( I_f = I_i + I_{\text{putty}} = 5.0 \times 10^{-4} + 2.0 \times 10^{-4} = 7.0 \times 10^{-4} \; \mathrm{kg \cdot m^2}\).
04

Apply Conservation of Angular Momentum

According to the conservation of angular momentum, the initial angular momentum \( L_i \) is equal to the final angular momentum \( L_f \). Therefore, \[ I_i \omega_i = I_f \omega_f \]where \( \omega_f \) is the final angular speed. Substitute the known values:\[ (5.0 \times 10^{-4})\cdot (4.7) = (7.0 \times 10^{-4}) \omega_f \]Solving for \( \omega_f \), we have:\[ \omega_f = \frac{(5.0 \times 10^{-4}) \times 4.7}{7.0 \times 10^{-4}} \approx 3.36 \; \mathrm{rad/s} \].
05

Conclusion

Thus, the angular speed of the record immediately after the putty sticks to it is approximately 3.36 rad/s, considering the conservation of angular momentum in the system.

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Key Concepts

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

Rotational Inertia
Rotational inertia, often referred to as the moment of inertia, is a fundamental concept when studying rotational motion. It represents how much an object resists changes to its rotational motion. Just as mass determines how much a linear motion resists acceleration, rotational inertia quantifies this for rotational systems.

Imagine a spinning vinyl record. It has rotational inertia because it requires effort to change its spinning speed. The rotational inertia depends on two factors:
  • The mass of the object
  • The distribution of that mass relative to the axis of rotation
For the vinyl record, its mass is distributed evenly along its radius. In our problem, the record's rotational inertia is initially given as \(5.0 \times 10^{-4} \; \mathrm{kg \cdot m^2}\). When additional mass is added to the system, like the wet putty sticking to the edge, the rotational inertia increases. More mass further from the axis intensifies resistance to angular acceleration, affecting the system's rotational characteristics.
Moment of Inertia
The moment of inertia is a key quantity in rotational dynamics. It acts as the rotational analog of mass in linear motion. The moment of inertia \(I\) can be calculated for various shapes depending on their mass distribution.

For instance, the formula we use when a point mass \(m\) is at a distance \(r\) from the axis is:\[ I = m r^2 \]In the exercise, the putty is treated as a point mass since it sticks at the edge of the record and affects the moment of inertia. Initially, the record had a moment of inertia \(I_i = 5.0 \times 10^{-4} \; \mathrm{kg \cdot m^2}\). After the putty sticks, it contributes additional inertia calculated by \(0.020 \; \mathrm{kg} \times (0.10 \; \mathrm{m})^2 = 2.0 \times 10^{-4} \; \mathrm{kg \cdot m^2}\).

The new total moment of inertia becomes \(I_f = I_i + I_{\text{putty}} = 7.0 \times 10^{-4} \; \mathrm{kg \cdot m^2}\). Understanding the moment of inertia helps us predict how the system's rotational behavior will change.
Angular Speed
Angular speed measures how fast an object rotates or revolves relative to another point. Usually denoted by \(\omega\), angular speed is similar to linear speed but in a rotational context. In our scenario, it's crucial for calculating the spinning rate of the vinyl record.

Initially, the record spins at an angular speed \(\omega_i = 4.7 \; \mathrm{rad/s}\). However, when the mass (the putty) is added, this speed changes due to the conservation of angular momentum. The principle states that if no external torque acts on the system, the total angular momentum remains constant. Hence,\[I_i \omega_i = I_f \omega_f\]Substituting the known initial values, we solve for \(\omega_f\) and find that:\[\omega_f = \frac{(5.0 \times 10^{-4}) \times 4.7}{7.0 \times 10^{-4}} \approx 3.36 \; \mathrm{rad/s}\].

This end result tells us how much slower the record spins after the putty sticks due to the increased moment of inertia. Understanding this shift in angular speed is essential in applying conservation laws to solve rotational dynamics problems.

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Most popular questions from this chapter

A wheel rotates clockwise about its central axis with an angular momentum of \(600 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). At time \(t=0\), a torque of magnitude \(50 \mathrm{~N} \cdot \mathrm{m}\) is applied to the wheel to reverse the rotation. At what time \(t\) is the angular speed zero?

A uniform wheel of mass \(10.0 \mathrm{~kg}\) and radius \(0.400 \mathrm{~m}\) is mounted rigidly on a massless axle through its center (Fig. 11-62). The radius of the axle is \(0.200 \mathrm{~m}\), and the rotational inertia of the wheel- axle combination about its central axis is \(0.600 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) The wheel is initially at rest at the top of a surface that is inclined at angle \(\theta=30.0^{\circ}\) with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by \(2.00 \mathrm{~m}\), what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

A \(140 \mathrm{~kg}\) hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of \(0.150 \mathrm{~m} / \mathrm{s}\). How much work must be done on the hoop to stop it?

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 disk with a rotational inertia of \(7.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) rotates like a merry-go-round while undergoing a variable torque given by \(\tau=(5.00+2.00 t) \mathrm{N} \cdot \mathrm{m}\). At time \(t=1.00 \mathrm{~s}\), its angular momentum is \(5.00 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\). What is its angular momentum at \(t=3.00 \mathrm{~s}\) ?

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