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Get Full Access to Sears And Zemansky's University Physics With Modern Physics - 13 Edition - Chapter 9 - Problem 87p
Get Full Access to Sears And Zemansky's University Physics With Modern Physics - 13 Edition - Chapter 9 - Problem 87p

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# Answer: Two metal disks, one with radius R1 = 2.50 cm and

ISBN: 9780321696861 89

## Solution for problem 87P Chapter 9

Sears and Zemansky's University Physics with Modern Physics | 13th Edition

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Sears and Zemansky's University Physics with Modern Physics | 13th Edition

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Problem 87P

Two metal disks, one with radius $$R_{1}=2.50 \mathrm{~cm}$$ and mass $$M_{1}=0.80 \mathrm{~kg}$$ and the other with radius $$R_{2}=5.00 \mathrm{~cm}$$ and mass $$M_{2}=1.60 \mathrm{~kg}$$,are welded together and mounted on a frictionless axis through their common center (Fig. P9.87).

(a) What is the total moment of inertia of the two disks?

(b) A light string is wrapped around the edge of the smaller disk, and a $$1.50-\mathrm{kg}$$ block is suspended from the free end of the string. If the block is released from rest at a distance of

$$2.00 \mathrm{~m}$$above the floor, what is its speed just before it strikes the floor?

(c) Repeat the calculation of part (b), this time with the string wrapped around the edge of the larger disk. In which case is the final speed of the block greater? Explain why this is so.

Equation Transcription:

Text Transcription:

R_1=2.50 cm

M_1=0.80 kg

R_2=5.00 cm

M_2=1.60 kg

1.50-kg

2.00 m

Step-by-Step Solution:

Step 1 of 9

Problem (a)

Radius of disk $$1 \mathrm{R}_{1}=2.50 \mathrm{~cm}$$ or $$2.5 \times 10^{-2} \mathrm{~m}$$

Radius of disk $$2 \mathrm{R}_{2}=5.00 \mathrm{~cm}$$ or $$5.0 \times 10^{-2} \mathrm{~m}$$

Mass of disk $$1 \mathrm{M}_{1}=0.8 \mathrm{~kg}$$

Mass of disk $$2 \mathrm{M}_{2}=1.6 \mathrm{~kg}$$

Step 2 of 9

Step 3 of 9

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