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The moment of inertia of the empty turntable is \(1.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\). With a constant torque of \(2.5 \mathrm{~N} \cdot \mathrm{m},\) the turntable-person system takes \(3.0 \mathrm{~s}\) to spin from rest to an angular speed of \(1.0 \mathrm{rad} / \mathrm{s} .\) What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. (a) \(2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (b) \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (c) \(7.5 \mathrm{~kg} \cdot \mathrm{m}^{2} ;\) (d) \(9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\).

Short Answer

Expert verified
The moment of inertia of the person about an axis through her center of mass is \(6.0 kg*m²\) (option b)

Step by step solution

01

Calculate the angular acceleration

Firstly, the angular acceleration can be calculated by using the equation \[α = Δω/Δt\] where Δω is the change in angular speed and Δt is the change in time. Here, the angular speed changes from 0 rad/s to 1 rad/s in 3 s. Therefore, the angular acceleration α is \[α = (1 rad/s - 0 rad/s) / 3 s = 1/3 rad/s²\]
02

Calculate the total moment of inertia

The torque τ exerted on the system of the turntable and the person is equal to the total moment of inertia \(I_tot\) of the system multiplied by the angular acceleration α. Therefore, the total moment of inertia \(I_tot\) can be calculated from \[I_tot = τ/α\] Given that the torque τ is 2.5 N*m and the angular acceleration α is 1/3 rad/s², the total moment of inertia \(I_tot\) is \[I_tot = 2.5 N*m /(1/3 rad/s²) = 7.5 kg*m²\]
03

Calculate the moment of inertia of the person

The total moment of inertia \(I_tot\) equals the sum of the moments of inertia of the turntable \(I_turntable\) and the person \(I_person\). Therefore, the moment of inertia of the person \(I_person\) can be calculated from \[I_person = I_tot - I_turntable\] Given that the total moment of inertia \(I_tot\) is 7.5 kg*m² and the moment of inertia of the turntable \(I_turntable\) is 1.5 kg*m², the moment of inertia of the person \(I_person\) is \[I_person = 7.5 kg*m² - 1.5 kg*m² = 6.0 kg*m²\]

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Key Concepts

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

Angular Acceleration Basics
Angular acceleration is a key concept in rotational dynamics, similar to how acceleration works in linear motion. It measures how quickly an object's rotational speed changes over time. This concept is directly linked to the change in angular velocity (\(\Delta \omega\)). In the given problem, we calculate the angular acceleration by finding the difference between the final and initial angular velocities and dividing by the change in time: \[\alpha = \frac{\Delta \omega}{\Delta t}\].
  • Initial Angular Speed (\(\omega_0\)): 0 rad/s
  • Final Angular Speed (\(\omega_f\)): 1 rad/s
  • Time Interval (\(\Delta t\)): 3 seconds
Using these values, the angular acceleration (\(\alpha\)) of the turntable is calculated as \(\alpha = \frac{1 \text{ rad/s} - 0 \text{ rad/s}}{3 \text{ s}} = \frac{1}{3} \text{ rad/s}^2\). Understanding angular acceleration helps us predict how fast the turntable will speed up.
Understanding Torque
Torque is crucial for understanding how rotational dynamics work. Think of torque as the rotational equivalent of linear force. It is what causes an object to rotate, depending on the force applied, the distance from the pivot point, and the angle of application. In simpler terms, torque is what makes the turntable spin.The formula for torque (\(\tau\)) is: \[\tau = r \times F \times \sin(\theta)\], but in this problem, it's known to be a constant value of 2.5 Nâ‹…m.
  • Force applied (\(F\)): related to the push or pull
  • Lever arm distance (\(r\)): how far from the pivot point the force is applied
The given torque value helps us calculate the total moment of inertia when combined with angular acceleration. By understanding torque, we can see how much rotational force is applied to make the turntable and person spin.
Basics of Rotational Dynamics
Rotational dynamics is the study of forces and torques and their effects on rotational motion. Just like how Newton's laws speak to objects in linear motion, similar principles exist for objects spinning, or rotating. These dynamics help us understand and predict how objects behave when they spin or turn.In the exercise, rotational dynamics helps in calculating the total moment of inertia by utilizing the known torque and angular acceleration. The formula connecting all these elements is: \[I_{\text{tot}} = \frac{\tau}{\alpha}\].
  • Total moment of inertia (\(I_{\text{tot}}\)): accounts for both the turntable and person's inertia
  • Helps predict the system's rotational behavior
By solving for the total inertia, we see how the system's resistance to rotational acceleration is calculated and how it includes both the person and turntable.
Turntable Rotations
A turntable involves rotation at a central pivot, much like a record player or a lazy Susan. The turntable in this exercise serves as a practical example of rotational dynamics in action.In our problem, it starts from rest and accelerates to a speed of 1.0 rad/s. The moment of inertia of the turntable, which is given as 1.5 kg⋅m², is important because it acts as the baseline rotational resistance without any added mass.
  • Moment of Inertia of the Turntable (\(I_{\text{turntable}}\)): 1.5 kgâ‹…m²
  • Acts as the base resistance to rotation
Incorporating a person on the turntable increases the moment of inertia due to added mass. Thus, understanding a turntable's moment of inertia allows us to determine the added inertia from the person, showcasing how different components in rotational systems interact.

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Most popular questions from this chapter

A woman with mass \(50 \mathrm{~kg}\) is standing on the rim of a large horizontal disk that is rotating at \(0.80 \mathrm{rev} / \mathrm{s}\) about an axis through its center. The disk has mass \(110 \mathrm{~kg}\) and radius \(4.0 \mathrm{~m} .\) Calculate the magnitude of the total angular momentum of the woman-disk system. (Assume that you can treat the woman as a point.)

A solid wood door \(1.00 \mathrm{~m}\) wide and \(2.00 \mathrm{~m}\) high is hinged along one side and has a total mass of \(40.0 \mathrm{~kg}\). Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass \(0.500 \mathrm{~kg}\), traveling perpendicular to the door at \(12.0 \mathrm{~m} / \mathrm{s}\) just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?

You live on a planet far from ours. Based on extensive communication with a physicist on earth, you have determined that all laws of physics on your planet are the same as ours and you have adopted the same units of seconds and meters as on earth. But you suspect that the value of \(g\), the acceleration of an object in free fall near the surface of your planet, is different from what it is on earth. To test this, you take a solid uniform cylinder and let it roll down an incline. The vertical height \(h\) of the top of the incline above the lower end of the incline can be varied. You measure the speed \(v_{\mathrm{cm}}\) of the center of mass of the cylinder when it reaches the bottom for various values of \(h .\) You plot \(v_{\mathrm{cm}}^{2}\) (in \(\mathrm{m}^{2} / \mathrm{s}^{2}\) ) versus \(h\) (in \(\mathrm{m}\) ) and find that your data lie close to a straight line with a slope of \(6.42 \mathrm{~m} / \mathrm{s}^{2}\). What is the value of \(g\) on your planet?

A \(2.00 \mathrm{~kg}\) textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is \(0.150 \mathrm{~m},\) to a hanging book with mass \(3.00 \mathrm{~kg} .\) The system is released from rest, and the books are observed to move \(1.20 \mathrm{~m}\) in \(0.800 \mathrm{~s}\) (a) What is the tension in each part of the cord? (b) What is the moment of inertia of the pulley about its rotation axis?

A thin uniform rod has a length of \(0.500 \mathrm{~m}\) and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of \(0.400 \mathrm{rad} / \mathrm{s}\) and a moment of inertia about the axis of \(3.00 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\). A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is \(0.160 \mathrm{~m} / \mathrm{s}\). The bug can be treated as a point mass. What is the mass of (a) the rod; (b) the bug?

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