Rolling Motion Question: Can someone help me out with this?

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1 Answer
Apr 1, 2018

Answer:

1.41 m, 1,85 revolutions

Explanation:

Let the speed of the center of mass of the ball at the bottom of the ramp be #v#. Then its angular velocity at that point is #omega = v/R#, and its kinetic energy is

#1/2 mv^2+1/2 I omega^2= 1/2 (m+I/R^2) v^2#

So, energy conservation gives

# 1/2 (m+I/R^2) v^2 = mg(h_1 - h_2)#

or

# v^2 = (2g(h_1-h_2))/(1+I/(mR^2)) = 10/7 g(h_1-h_2) = 10/7 times 10" m"\ "s"^-2 times 0.35" m" = 5" m"^2"s"^2#

(here we have used the fact that #I=2/5mR^2# for a solid sphere.) So

#v = sqrt(5)" m/s"#

The time the ball takes to reach the floor from this point on is given by

#t = sqrt((2h_2)/g) = sqrt((2 times 2" m")/(10" m"\ "s"^2)) = sqrt(2/5)" s"#

Thus, the horizontal distance the ball travels before hitting the floor is

# sqrt(5) " m/s" times sqrt(2/5)" s" = sqrt(2)" m"~~1.41" m"#

The angle the ball revolves through in time #t# is given by

#theta = omega times t = v/R times t ~~ 1.41/0.12 " rad" #

Thus, the number of revolutions is

#theta/(2 pi) ~~ 1.85#