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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

The Yo-yo. A yo-yo is made from two uniform disks, each with mass \(m\) and radius \(R\), connected by a light axle of radius \(b\). A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass \(5.00 \mathrm{~kg}\) and radius \(0.120 \mathrm{~m} .\) For each the translational speed of the center of mass is \(4.00 \mathrm{~m} / \mathrm{s}\). Sphere \(A\) is a uniform solid sphere and sphere \(B\) is a thin-walled, hollow sphere. How much work, in joules, must be done on each sphere to bring it to rest? For which sphere is a greater magnitude of work required? Explain. (The spheres continue to roll without slipping as they slow down.

If the body's center of mass were not placed on the rotational axis of the turntable, how would the person's measured moment of inertia compare to the moment of inertia for rotation about the center of mass? (a) The measured moment of inertia would be too large; (b) the measured moment of inertia would be too small; (c) the two moments of inertia would be the same; (d) it depends on where the body's center of mass is placed relative to the center of the turntable.

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to \(80.0 \mathrm{~N}\) is applied to the rim of the wheel. The wheel has radius \(0.120 \mathrm{~m}\) Starting from rest, the wheel has an angular speed of \(12.0 \mathrm{rev} / \mathrm{s}\) after \(2.00 \mathrm{~s}\). What is the moment of inertia of the wheel?

A cord is wrapped around the rim of a solid uniform wheel \(0.250 \mathrm{~m}\) in radius and of mass \(9.20 \mathrm{~kg} .\) A steady horizontal pull of \(40.0 \mathrm{~N}\) to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axle exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

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