Question

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

Answer 1

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`