I = 1/2MR^2 is rolling with a translational speed v along a horizontal surface without slipping. The work required to stop the disk is A. 1/2 mv^2 B. 1/2mv^2 + 1/2Iw^2 C. 1/2Iw^2 D. 1/2mv^2 - 1/2Iw^2

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asked Apr 12, 2024 120k views

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I = 1/2MR^2 is rolling with a translational speed v along a horizontal surface without slipping. The work required to stop the disk is

A. 1/2 mv^2
B. 1/2mv^2 + 1/2Iw^2
C. 1/2Iw^2
D. 1/2mv^2 - 1/2Iw^2

Physics
college

Jon Onstott asked

by Jon Onstott

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1 Answer

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Answer:

B. 1/2mv^2 + 1/2Iw^2

Step-by-step explanation:

Sebnukem answered Apr 16, 2024

by Sebnukem

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