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A thin spherical shell has a radius of 1.88 m. An applied torque of 960 Nm imparts an angular acceleration equal to 6.23rad/s2 about an axis through the center of the shell. Calculate (a) the rotational inertia of the shell about the axis of rotation and (b) the mass of the shell.

Short Answer

Expert verified
The rotational inertia of the shell is I=τ/α and the mass of the shell is m=I/(2/3r2).

Step by step solution

01

Find the Rotational Inertia

Using the formula for rotational inertia I=τ/α, insert the provided values: I=960 Nm/6.23rad/s2. Calculate the solution to find I.
02

Calculate the Mass

Rearrange the formula I=2/3mr2 to solve for m, giving m=I/(2/3r2). Insert the known values of I and r to calculate the mass.

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Key Concepts

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

Torque
Torque is a crucial concept in understanding rotational motion. It is often described as the rotational equivalent of linear force.
When you apply a force to rotate an object around an axis, you're essentially applying torque. The amount of torque applied depends on three factors:
  • The magnitude of the force.
  • The distance from the axis to where the force is applied (lever arm).
  • The angle between the force direction and the line of action on the lever arm.
For this problem, the provided torque is measured in newton-meters (N·m), which tells us the rotational force applied to the spherical shell.
In mathematical terms, you calculate torque using the equation:τ=Frsin(θ)where F is the force applied, r is the distance to the rotation axis, and θ is the angle involved. When the force is applied perpendicular to the lever arm, θ=90°, and sin(θ)=1.
This makes torque simply the product of force and distance, τ=Fr, assuming the most effective scenario, which is the case presented here: τ=960Nm.
Understanding how torque influences rotational motion is essential for solving any rotational dynamics problems like the one presented.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time. It describes how fast an object changes its speed of rotation.
This concept is quite similar to linear acceleration but applies to rotational motion. In the exercise, the angular acceleration is provided as 6.23rad/s2. This value signifies how much the rotational speed of the spherical shell changes per second.
Angular acceleration is generally denoted by α (alpha). The relationship between torque (τ), rotational inertia (I), and angular acceleration is given by the equation:τ=IαThis formula tells us that the torque applied to an object is directly proportional to its angular acceleration and its moment of inertia. Given that the spherical shell in the problem has a specific angular acceleration, it means that the torque is sufficient to change its rotational speed by 6.23rad/s2.
Grasping angular acceleration is vital to understanding how different factors affect the rotational motion of objects.
Moment of Inertia
The moment of inertia, often denoted as I, is a property that quantifies how difficult it is to change the rotational speed of an object.
Think of it as the rotational equivalent of mass in linear motion. The larger the moment of inertia, the more torque you need to achieve a certain angular acceleration.
In this problem, using the relationship I=τ/α, we can calculate the moment of inertia of the shell by dividing the torque by the angular acceleration:I=960Nm6.23rad/s2Upon solving this, you'll find the exact rotational inertia for the given situation. For homogenous spherical shells, the formula linking mass and moment of inertia is typically:I=23mr2Rearranging this formula allows us to determine the mass of the shell once I is known. The moment of inertia is crucial when dealing with problems involving rotational dynamics, offering insight into how an object's mass is distributed relative to the rotation axis. Understanding this concept is essential for solving exercises related to rotational motion.

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Most popular questions from this chapter

A yo-yo (see Sample Problem 913 ) has a rotational inertia of 950 gcm2 and a mass of 120 g. Its axle radius is 3.20 mm and its string is 134 cm long. The yo-yo rolls from rest down to the end of the string. ( a ) What is its acceleration? (b) How long does it take to reach the end of the string? (c) If the yo-yo "sleeps" at the bottom of the string in pure rotary motion, what is its angular speed, in rev/s? (d) Repeat (c), but this time assume that the yo-yo was thrown down with an initial speed of 1.30 m/s

Two identical blocks, each of mass M, are connected by a light string over a frictionless pulley of radius R and rotational inertia I (Fig. 9-55). The string does not slip on the pulley, and it is not known whether or not there is friction between the plane and the sliding block. When this system is released, it is found that the pulley turns through an angle θ in time t and the acceleration of the blocks is constant. (a) What is the angular acceleration of the pulley? ( b ) What is the acceleration of the two blocks? ( c ) What are the tensions in the upper and lower sections of the string? All answers are to be expressed in terms of M,I,R,θ,g, and t.

What is the torque about the origin on a particle located at x=1.5 m,y=2.0 m,z=1.6 m due to a force F= (3.5 N)i^(2.4 N)j^+(4.3 N)k^? Express your result in unit vector notation.

Vector a has magnitude 3.20 units and lies in the yz plane 63.0 from the +y axis with a positive z component. Vector b has magnitude 1.40 units and lies in the xz plane 48.0 from the +x axis with a positive z component. Find a×b.

Two vectors r and s lie in the xy plane. Their magnitudes are r=4.5 units and s=7.3 units. Their directions are, respectively, 320 and 85 measured counterclockwise from the positive x axis. Find the magnitude and the direction of r×s

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