NEXT Question 9 Your answer is partially correct. Try again. Starting from rest,

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NEXT Question 9 Your answer is partially correct. Try again. Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.9 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom? (a) Numben (b) Number-de- enUnitsto Units m/s Click if you would like to Show Work for this question: Open Show Work SHOW HINT LINK TO TEXT Link to Sample Problem Question Attempts: 3 of 5 used SAVE FOR LATER Copyright © 2000-2015 by John Wiley & Sons, Inc. or related companies. All rights reserved ntolicy | 2000-2015 Jahn v ilev sens Inc. All Rights Reserved. A Division of lohnwilu&sons; inc Version 4.13.2.2

Explanation / Answer

a)

angular speed

W=V/r

Moment of inerita of basketball (sphere)

I=(2/3)mr2

By Conservation of energy

(1/2)mV2+(1/2)IW2=mgh

(1/2)mV2+(1/2)[(2/3)mr2)](V/r)2=mgh

h=5V2/6g =5*6.92/6*9.8

h=4.05 m

b)

Moment of inertia

I=(1/2)mr2

By conservation of energy

(1/2)mV2+(1/2)IW2=mgh

(1/2)mV2+(1/2)[(1/2)mr2)](V/r)2=mgh

=>V=sqrt[4*g*h/3]=sqrt[4*9.8*4.05/3]

V=7.27 m/s

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