Giancoli's Physics: Principles with Applications, 7th Edition
8
Rotational Motion
Change chapter8-1: Angular Quantities
8-2 and 8-3: Constant Angular Acceleration; Rolling
8-4: Torque
8-5 and 8-6: Rotational Dynamics
8-7: Rotational Kinetic Energy
8-8: Angular Momentum
8-9: Angular Quantities as Vectors
Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 6
Q
A child rolls a ball on a level floor 3.5 m to another child. If the ball makes 12.0 revolutions, what is its diameter?
A
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VIDEO TRANSCRIPT
This is Giancoli Answers with Mr. Dychko. The total distance this ball travels when it rolls twelve times is 3.5 meters and each revolution of the ball, it will travel a distance of the circumference of the ball. And so the total distance covered then is the number of revolutions times the circumference—number of meters per revolution— and the circumference is π times diameter. So we can solve for diameter by dividing l and this side by number of revolutions times π. So diameter is l over number of revolutions times π so that's 3.5 meters divided by 12.0 and divide by π and that gives 9.3 centimeters must be the diameter of the ball.
COMMENTS
By rafih6 on Thu, 11/19/2015 - 12:24 AM
Nice dude
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