Conservation of energy, Rolling without slipping.


by quantumlolz
Tags: conservation, energy, rolling, slipping
quantumlolz
quantumlolz is offline
#1
Apr11-10, 09:48 PM
P: 8
1. The problem statement, all variables and given/known data

A bowling ball is rolling without slipping up an inclined plane. As it passes a point O it has a speed of 2.00 m/s up the plane. It reaches a vertical height h above O before momentarily stopping and rolling back down. Determine the value of h. The moment of inertia of a solid sphere of mass M and radius R is [tex]I = \frac{2MR^2}{5}[/tex]

2. Relevant equations

Translational kinetic energy [tex] \frac{mv^2}{2}[/tex]
Rotational kinetic energy [tex] \frac {I\omega^2}{2} [/tex]
Rolling without slipping [tex] v = r\omega [/tex]
Gravitational potential energy [tex] U = mgy [/tex]
Constant acceleration formulae maybe?

3. The attempt at a solution

Equating Total (rotational and translational) kinetic energy and potential energy [tex] \frac{Mv^2}{2} + \frac {I\omega^2}{2} = Mgy [/tex] as all the energy has been converted from kinetic to potential when the ball stops (i.e. at the height we're trying to find)

Inserting the expression for moment of inertia the rotational kinetic energy is [tex] \frac{MR^2\omega^2}{5} [/tex] but for rolling without slipping this is just [tex] \frac{Mv^2}{5} [/tex] so total kinetic energy becomes [tex] T = \frac{Mv^2}{2} + \frac{Mv^2}{5} = \frac {7Mv^2}{10} [/tex]

So [tex] \frac{7Mv^2}{10} = Mgy [/tex] and we can cancel the Ms to get [tex] \frac {7v^2}{10} = gy [/tex]

At some value of y (lets call it d), v = 2.00 m/s so plugging in the numbers [tex] d = \frac{7v^2}{10g} [/tex] = 0.29m

Then I'm not sure what to do next. I was thinking we've got an initial velocity (2 m/s), a final velocity (0 m/s), an acceleration (g), and we want the vertical displacement, h, from O to give a final velocity of 0 m/s so we can then apply constant acceleration equation [tex] v^2 = u^2 + 2gh [/tex], with v = 2, u = 0 and solving this gives [tex] h = \frac{v^2}{2g} [/tex] but this comes to h = 0.20m, the quoted answer is h = 0.29 m.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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ehild
ehild is offline
#2
Apr11-10, 11:42 PM
HW Helper
Thanks
P: 9,817
You got the correct result, h=0.29 m, only you did not notice.

You can take the potential equal to zero at any point, here a good choice is at point O.

At point O the ball has vo=2m/s velocity and it rolls upward. You calculated the KE with this velocity, correctly. This KE transforms to potential energy h height above O, where the potential energy with respect to O, is mgh. As you wrote,

[tex]
d = \frac{7v^2}{10g} =0.29 m
[/tex]

But d is the height h which was the question. You got it!

ehild
quantumlolz
quantumlolz is offline
#3
Apr12-10, 07:28 AM
P: 8
Quote Quote by ehild View Post
You got the correct result, h=0.29 m, only you did not notice.

You can take the potential equal to zero at any point, here a good choice is at point O.

At point O the ball has vo=2m/s velocity and it rolls upward. You calculated the KE with this velocity, correctly. This KE transforms to potential energy h height above O, where the potential energy with respect to O, is mgh. As you wrote,

[tex]
d = \frac{7v^2}{10g} =0.29 m
[/tex]

But d is the height h which was the question. You got it!

ehild
Ah cool, I did spot that I got the right number when I was working it out, but wasnt sure why - turns out I'd completely forgotten that I can choose where the potential is zero. Thanks :)


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