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Ch 09: Rotation of Rigid Bodies
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 15th EditionCh 09: Rotation of Rigid BodiesProblem 41
Chapter 9, Problem 41

A uniform sphere with mass 28.028.0 kg and radius 0.3800.380 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 236236 J, what is the tangential velocity of a point on the rim of the sphere?

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Step 1: Recall the formula for rotational kinetic energy, which is given by \( KE = \frac{1}{2} I \omega^2 \), where \( KE \) is the rotational kinetic energy, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
Step 2: For a uniform sphere rotating about a diameter, the moment of inertia \( I \) is \( \frac{2}{5} m r^2 \), where \( m \) is the mass of the sphere and \( r \) is its radius. Substitute the given values \( m = 28.0 \ \text{kg} \) and \( r = 0.380 \ \text{m} \) into this formula to calculate \( I \).
Step 3: Rearrange the rotational kinetic energy formula to solve for \( \omega \): \( \omega = \sqrt{\frac{2 KE}{I}} \). Substitute the given \( KE = 236 \ \text{J} \) and the calculated \( I \) from Step 2 into this equation to find \( \omega \).
Step 4: The tangential velocity \( v_t \) of a point on the rim of the sphere is related to the angular velocity \( \omega \) by the formula \( v_t = r \omega \). Use the radius \( r = 0.380 \ \text{m} \) and the calculated \( \omega \) from Step 3 to find \( v_t \).
Step 5: Combine all the results from the previous steps to express \( v_t \) in terms of the given quantities. This will give the tangential velocity of a point on the rim of the sphere.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Angular Velocity

Angular velocity is a measure of how quickly an object rotates around an axis, expressed in radians per second. For a rotating object, it relates to the linear velocity of points on the object’s surface through the equation v = ωr, where v is the tangential velocity, ω is the angular velocity, and r is the radius. Understanding angular velocity is crucial for determining the motion of points on a rotating sphere.
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Kinetic Energy of Rotation

The kinetic energy of a rotating object is given by the formula KE = 0.5 I ω², where I is the moment of inertia and ω is the angular velocity. For a uniform sphere, the moment of inertia is I = (2/5)mr², where m is the mass and r is the radius. This concept is essential for relating the sphere's kinetic energy to its rotational motion and ultimately finding the tangential velocity.
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Tangential Velocity

Tangential velocity is the linear speed of a point on the circumference of a rotating object, calculated as v = ωr. It represents how fast a point moves along its circular path and is directly proportional to both the angular velocity and the radius of the object. In this problem, finding the tangential velocity of a point on the rim of the sphere requires understanding its relationship with angular velocity and radius.
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Related Practice
Textbook Question

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Textbook Question

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A wheel is turning about an axis through its center with constant angular acceleration. Starting from rest, at t = 0, the wheel turns through 8.20 revolutions in 12.0 s. At t = 12.0 s the kinetic energy of the wheel is 36.0 J. For an axis through its center, what is the moment of inertia of the wheel?

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Textbook Question

Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M and radius R about an axis perpendicular to the hoop's plane at an edge.

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Textbook Question

An airplane propeller is 2.08 m in length (from tip to tip) with mass 117 kg and is rotating at 2400 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod.

(a) What is its rotational kinetic energy?

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Textbook Question

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