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Chapter 8: Problem 61

A solid, horizontal cylinder of mass \(10.0 \mathrm{~kg}\) and radius \(1.00 \mathrm{~m}\) rotates with an angular speed of \(7.00 \mathrm{rad} / \mathrm{s}\) about a fixed vertical axis through its center. A \(0.250-\mathrm{kg}\) piece of putty is dropped vertically onto the cylinder at a point \(0.900 \mathrm{~m}\) from the center of rotation and sticks to the cylinder. Determine the final angular speed of the system.

Short Answer

Expert verified
The final angular speed of the system can be found by using the law of conservation of angular momentum to relate the initial angular momentum of the cylinder to the final angular momentum of the system.

Step by step solution

01

Calculate the Initial Angular Momentum

First, determine the moment of inertia of the cylinder, \(I_C\), using the formula \(I = 0.5m r^2\), where \(r\) is the radius and \(m\) is the mass of the cylinder. Then, compute the initial angular momentum, \(L_i\), which equals to the moment of inertia of the cylinder multiplied by its angular speed \(L_i = I_C \omega\).
02

Get the Final Moment of Inertia

Next, calculate the moment of inertia of the putty, \(I_P\), using the formula \(I = m r^2\), where \(m\) is the mass of the putty and \(r\) is the distance from the axis of rotation. Sum the moments of inertia of the cylinder and the putty to find the total final moment of inertia \(I_F = I_C + I_P\).
03

Find the Final Angular Speed

Finally, solve for the final angular speed, \(\omega'\), using the law of conservation of angular momentum (\(L_i = L_f\)), where \(L_f = I_F \omega'\). Rearrange the equation to solve for \(\omega'\), giving \(\omega' = L_i / I_F\).

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Key Concepts

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

Moment of Inertia
Moment of inertia, often referred to as rotational inertia, is a measure of an object's resistance to changes in its rotation rate. It plays the same role in rotational motion as mass does in linear motion. Just as a larger mass requires more force to accelerate, a larger moment of inertia requires more torque to change its rotational speed.

For different shapes and mass distributions, the moment of inertia has different expressions. For a solid cylinder like the one in our exercise, the moment of inertia \(I_C\) is given by the formula \(I_C = 0.5mr^2\), where \(m\) is the mass of the cylinder and \(r\) is its radius. This is crucial for calculating how much the cylinder will slow down when an additional mass, such as a piece of putty, is added at a certain radius from its center of rotation.
Conservation of Angular Momentum
The principle of conservation of angular momentum states that if no external torque acts on a system, the total angular momentum of that system remains constant. Angular momentum is the product of the moment of inertia and the angular speed of an object.

In our example, when the putty drops onto the cylinder and sticks to it, the system's total angular momentum must remain the same because there's no external torque. However, the moment of inertia changes because the putty has its own moment of inertia and contributes to the total inertia of the system. Since the angular momentum must stay the same and the moment of inertia increases, the final angular speed of the cylinder decreases. This conservation law allows us to solve many rotational dynamics problems by setting the initial angular momentum equal to the final angular momentum, \(L_i = L_f\).
Angular Speed

Understanding Angular Speed

Angular speed, \(\omega\), is a scalar measure of the rotation rate of an object. It indicates how many radians an object rotates through per unit of time. In simple terms, it tells us how fast an object is spinning.

When an object like our cylinder is rotating without additional forces or masses being added, its angular speed remains constant. However, when the mass distribution changes, as when the putty is added, the angular speed can change. To find the new angular speed after the putty sticks to the cylinder, we use the conservation of angular momentum. Given the initial angular momentum and the new moment of inertia, \(I_F\), the new angular speed \(\omega'\) can be calculated. As the moment of inertia increases, the angular speed must decrease for the total angular momentum to be the same, reflecting an inverse relationship between the two quantities in a closed system.

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Most popular questions from this chapter

A \(40.0\) -kg child stands at one end of a \(70.0\) -kg boat that is \(4.00 \mathrm{~m}\) long (Fig. \(\mathrm{P} 8.69\) ). The boat is initially \(3.00 \mathrm{~m}\) from the pier. The child notices a turtle on a rock beyond the far end of the boat and proceeds to walk to that end to catch the turtle. (a) Neglecting friction between the boat and water, describe the motion of the system (child plus boat). (b) Where will the child be relative to the pier when he reaches the far end of the boat? (c) Will he catch the turtle? (Assume that he can reach out \(1.00 \mathrm{~m}\) from the end of the boat.)

An Atwood's machine consists of blocks of masses \(m_{1}=10.0 \mathrm{~kg}\) and \(m_{2}=20.0 \mathrm{~kg}\) attached by a cord running over a pulley as in Figure \(\mathrm{P} 8.40 .\) The pulley is a solid cylinder with mass \(M=8.00 \mathrm{~kg}\) and radius \(r=0.200 \mathrm{~m}\) The block of mass \(m_{2}\) is allowed to drop, and the cord turns the pulley without slipping. (a) Why must the tension \(T_{2}\) be greater than the tension \(T_{1} ?\) (b) What is the acceleration of the system, assuming the pulley axis is frictionless? (c) Find the tensions \(T_{1}\) and \(T_{2}\).

An airliner lands with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\). Each wheel of the plane has a radius of \(1.25 \mathrm{~m}\) and a moment of inertia of \(110 \mathrm{~kg} \cdot \mathrm{m}^{2}\). At touchdown, the wheels begin to spin under the action of friction. Each wheel supports a weight of \(1.40 \times 10^{4} \mathrm{~N}\), and the wheels attain their angular speed in \(0.480 \mathrm{~s}\) while rolling without slipping. What is the coefficient of kinetic friction between the wheels and the runway? Assume that the speed of the plane is constant.

A playground merry-go-round of radius \(2.00 \mathrm{~m}\) has a moment of inertia \(I=275 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and is rotating about a frictionless vertical axle. As a child of mass \(25.0 \mathrm{~kg}\) stands at a distance of \(1.00 \mathrm{~m}\) from the axle, the system (merrygo-round and child) rotates at the rate of \(14.0 \mathrm{rev} / \mathrm{min}\). The child then proceeds to walk toward the edge of the merry-go-round. What is the angular speed of the system when the child reaches the edge?

A constant torque of \(25.0 \mathrm{~N} \cdot \mathrm{m}\) is applied to a grindstone whose moment of inertia is \(0.130 \mathrm{~kg} \cdot \mathrm{m}^{2}\). Using energy principles and neglecting friction, find the angular speed after the grindstone has made \(15.0\) revolutions, Hint: The angular equivalent of \(W_{\mathrm{net}}=F \Delta x=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}\) is \(W_{\text {net }}\) \(=\tau \Delta \theta=\frac{1}{2} I \omega_{f}^{2}-\frac{1}{2} J \omega_{i}^{2} .\) You should convince yourself that this relationship is correct.
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