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Chapter 9: Problem 38

A uniform sphere with mass 28.0 \(\mathrm{kg}\) and radius 0.380 \(\mathrm{m}\) is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is \(176 \mathrm{J},\) what is the tangential velocity of a point on the rim of the sphere?

Short Answer

Expert verified
The tangential velocity is approximately 5.01 m/s.

Step by step solution

01

Identify what is given

We know the mass \( m = 28.0 \, \mathrm{kg} \), radius \( r = 0.380 \, \mathrm{m} \), and kinetic energy \( KE = 176 \, \mathrm{J} \) of the sphere. We are asked to find the tangential velocity of a point on the rim of the sphere.
02

Recall the formula for rotational kinetic energy

The rotational kinetic energy \( KE \) for a rotating object is given by: \[ KE = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
03

Find the moment of inertia for a solid sphere

For a solid sphere rotating about an axis through its center, the moment of inertia \( I \) is: \[ I = \frac{2}{5} m r^2 \] Substitute \( m = 28.0 \, \mathrm{kg} \) and \( r = 0.380 \, \mathrm{m} \) to find \( I \): \[ I = \frac{2}{5}(28.0)(0.380)^2\] Calculate to find \( I \approx 2.0256 \, \mathrm{kg} \, \mathrm{m}^2 \).
04

Solve for angular velocity \( \omega \)

Now solve the equation \( KE = \frac{1}{2} I \omega^2 \) for \( \omega \): \[ 176 = \frac{1}{2} (2.0256) \omega^2 \] \[ \omega^2 = \frac{176}{1.0128} \] \[ \omega^2 \approx 173.8 \] \[ \omega \approx 13.18 \, \mathrm{rad/s} \].
05

Calculate the tangential velocity \( v_t \)

The tangential velocity \( v_t \) at the rim of the sphere is given by: \[ v_t = r \omega \] Substitute \( r = 0.380 \, \mathrm{m} \) and \( \omega \approx 13.18 \, \mathrm{rad/s} \) to find \( v_t \): \[ v_t = 0.380 \times 13.18 \] \[ v_t \approx 5.01 \, \mathrm{m/s} \].

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Key Concepts

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

Understanding Moment of Inertia
Moment of inertia, often symbolized as \( I \), measures an object's resistance to changes in its rotational motion. It is somewhat analogous to mass in linear motion but specifically for rotation.
Here are some key points to understand about moment of inertia:
  • It depends not only on the mass of the object but also on how that mass is distributed relative to the axis of rotation.
  • For simple objects and geometries, like a solid sphere, we have specific formulas to calculate \( I \). For example, the formula for a solid sphere is \( I = \frac{2}{5} m r^2 \).
  • The greater the moment of inertia, the more rotational kinetic energy is required to change its angular velocity.
Using the given formula, plugging in the mass \( m = 28.0 \, \mathrm{kg}\) and the radius \( r = 0.380 \, \mathrm{m}\), we found that \( I \) for the sphere is approximately \( 2.0256 \, \mathrm{kg} \, \mathrm{m}^2 \). This value indicates the resistance of the sphere to angular acceleration around its axis.
What is Angular Velocity?
Angular velocity \( \omega \) describes how fast an object rotates or spins around an axis. It is a vector quantity, which means it has both magnitude and direction. However, in most basic problems, it is sufficient to focus on the magnitude.
Some essential aspects of angular velocity are:
  • It is measured in radians per second (rad/s), providing an angular displacement (in radians) per unit time.
  • Angular velocity is closely linked with the rotational kinetic energy of an object. Specifically, kinetic energy, \( KE = \frac{1}{2} I \omega^2 \), shows how \( \omega \) affects the energy needed for the rotation.
  • Changing the angular velocity of an object implies either speeding it up or slowing it down in its rotational motion.
In our example, we found the angular velocity of the sphere to be approximately \( 13.18 \, \mathrm{rad/s} \) after considering the given rotational kinetic energy.
Exploring Tangential Velocity
Tangential velocity \( v_t \) is the linear speed of a point located at a particular distance from the axis of rotation. Imagine a point on a spinning wheel; the path it traces as it spins defines its tangential movement.
Here are key points about tangential velocity:
  • It exists at any point that moves along the circumference of a circle centered on the axis of rotation.
  • Tangential velocity is calculated as \( v_t = r \omega \), where \( r \) is the distance from the rotation axis (radius in the case of a sphere) and \( \omega \) is the angular velocity.
  • An object's tangential velocity increases with either a greater \( \omega \) or a larger radius \( r \).
In the situation we considered, we calculated the tangential velocity of a point on the rim as approximately \( 5.01 \, \mathrm{m/s} \). This shows how the rotational motion translates into linear motion at the surface of the sphere.

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Most popular questions from this chapter

An electric turntable 0.750 \(\mathrm{m}\) in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 \(\mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of 0.900 \(\mathrm{rev} / \mathrm{s}^{2}\) . (a) Compute the angular velocity of the turntable after 0.200 \(\mathrm{s}\) . (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200\) s? (d) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{s} ?\)

A flywheel has angular acceleration \(\alpha_{z}(t)=$$8.60 \mathrm{rad} / \mathrm{s}^{2}-\left(2.30 \mathrm{rad} / \mathrm{s}^{3}\right) t,\) where counterclockwise rotation is positive. (a) If the flywheel is at rest at \(t=0,\) what is its angular velocity at 5.00 s? (b) Through what angle (in radians) does the flywheel turn in the time interval from \(t=0\) to \(t=5.00\) s?

A light, flexible rope is wrapped several times around a hollow cylinder, with a weight of 40.0 \(\mathrm{N}\) and a radius of 0.25 \(\mathrm{m}\), that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligible moment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force \(P\) for a distance of \(5.00 \mathrm{m},\) at which point the end of the rope is moving at 6.00 \(\mathrm{m} / \mathrm{s} .\) If the rope does not slip on the cylinder, what is the value of \(P ?\)

A disk of radius 25.0 \(\mathrm{cm}\) cm is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. P9.65). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is 1.80 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) Find \(A .\) (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .\) .

A turntable rotates with a constant 2.25 \(\mathrm{rad} / \mathrm{s}^{2}\) angular acceleration. After 4.00 s it has rotated through an angle of 60.0 rad. What was the angular velocity of the wheel at the beginning of the 4.00 -s interval?
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