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a) $a_c=1.42m/s^2$
b) $F_c=35.5N$

Giancoli 6th Edition, Chapter 5, Problem 1

(2:41)

Chapter 5, Problem 1 is solved.

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Transcript for this Giancoli solution

So let's figure out the centripetal acceleration of this child on the merry-go-round. So we'll draw the merry-go-round here, and then put our information onto this picture. We have that the radius is one point one zero meters and the speed of the child (the tangential velocity) is one point two five meters per second. We have the formula for centripetal acceleration (we'll do part a first), the formula is 'a' equals 'v' squared over 'r', and we'll substitute in these numbers: one point two five meters per second all squared divided by one point one zero meters and this gives us one point four two zero four five meters per second squared and we're going to keep lots of digits just because we're going to use this number in part 'b'. And the final answer for part 'a' is one point four two meters per second squared. For part 'b' we'll find the centripetal force, which is mass times centripetal acceleration. And this is twenty-five point zero kilograms times our answer from part 'a', one point four two oh four five meters per second squared, and we get thirty-five point five one one Newtons, which with three sig. Figs. Is thirty-five point five Newtons. For the fifth edition you'll have one point two meters here for the radius and you'll have one point three five meters per second and you'll substitute those numbers in. So it's one point three five here and it's one point two zero meters here. Your answer will be one point five two meters per second squared here, and substituting that here will give you an answer for a centripetal force of thirty-eight point zero Newtons for the 5th Edition.

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