MasteringPhysics: Homework 6-1 - Mozilla Firefox A https://session.masteringphys
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MasteringPhysics: Homework 6-1 - Mozilla Firefox A https://session.masteringphysics.com/myct/itemView?offset=next&assignmentProblemID;=86355354 - O X 133% *** D * = Resources - 9 « previous | 14 of 14 | return to assignment Homework 6-1 Enhanced EOC: Problem 12.34 Enhanced EOC: Problem 12.34 Part A What is the sphere's angular velocity at the bottom of the incline? Express your answer with the appropriate units. · Hints An 8.00-cm-diameter, 370 g sphere is released from rest at the top of a 1.50-m-long, 18.0° incline. It rolls, without slipping, to the bottom. You may want to review ( pages 334 - 337). For help with math skills, you may want to review: Solving Algebraic Equations Conversion Between m and cm Conversion Between mg. g. kg PHLÅ 6 - 0 E ? | Value Units Submit My Answers Give Up Part B What fraction of its kinetic energy is rotational? · Hints VA A2 0 ? Submit My Answers Give UpExplanation / Answer
A)
I assume it is a solid sphere.
Height of incline, h = 1.50sin18° = 0.4635 m
Loss of potential energy = gain of kinetic energy
=> mgh = (1/2) mv^2 + (1/2) I ^2
=> mgh = (1/2) mv^2 + (1/2) * (2/5) mR^2^2
=> gh = (1/2) v^2 + (1/5) v^2 = (7/10) v^2
=> v = [(10/7)gh] = [(10/7)9.8 * 0.4635]
=> v = 2.547 m/s
=> = (v/R) = 2.547 / 0.04 = 63.69 rad/s.
B)
Total KE = (7/10) mv^2
Rotational KE = (1/5)mv^2
Rotational KE / Total KE
= (1/5) * (10/7) = 2/7.
Hope this helps :)