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PostPosted: Mon, 23 Apr 2007 04:33:54 UTC 
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While exploring a castle, Exena the Exterminator is spotted by a dragon who chases her down a hallway. Exena runs into a room and attempts to swing the heavy door shut before the dragon gets her. The door is initially perpendicular to the wall, so it must be turned through 90 degrees to close. The door has a height of a and a width of b , and its weight is w . You can ignore the friction at the hinges.

If Exena applies a force of F at the edge of the door and perpendicular to it, how much time does it take her to close the door?

Use g as acceleration due to gravity.

This question seems simple, however im not too sure how to go about it. I've done a few simpler angular velocity/acceleration questions and they were straightforward... Any help would be greatly appreciated :)


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PostPosted: Mon, 23 Apr 2007 14:20:08 UTC 
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bally wrote:
While exploring a castle, Exena the Exterminator is spotted by a dragon who chases her down a hallway. Exena runs into a room and attempts to swing the heavy door shut before the dragon gets her. The door is initially perpendicular to the wall, so it must be turned through 90 degrees to close. The door has a height of a and a width of b , and its weight is w . You can ignore the friction at the hinges.

If Exena applies a force of F at the edge of the door and perpendicular to it, how much time does it take her to close the door?

Use g as acceleration due to gravity.

This question seems simple, however im not too sure how to go about it. I've done a few simpler angular velocity/acceleration questions and they were straightforward... Any help would be greatly appreciated :)


Start by finding the mass of door, hence the moment of inertia of the door about suitable axis, modelling the door as a uniform lamina of prescribed dimensions.


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PostPosted: Mon, 23 Apr 2007 14:22:37 UTC 
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So Mass of the door is w/g. How do I get the moment of inertia? It will obviously be around the y axis...


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PostPosted: Mon, 23 Apr 2007 21:21:29 UTC 
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here's a link to a table with different moments of inertia ... one for your problem is on it.

$ \tau_{net} = I\alpha

$ r \times F = I\alpha

$ \frac{r \times F}{I} = \alpha

once you figure the constant angular acceleration, use the following kinematics equation to solve for t ...

$ \Delta \theta = \omega_o t + \frac{1}{2} \alpha t^2


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PostPosted: Tue, 24 Apr 2007 11:21:27 UTC 
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Well, so far I have got:

Moment of Inertia = 1/3 * (w.g) . a^2

What do I do next? Since I can't find out the numerical value of inertia since the values are arbitrary.


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PostPosted: Tue, 24 Apr 2007 13:38:19 UTC 
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you've never solved a physics problem with given quantities in terms of variables?

in any case, your expression for the moment of inertia is incorrect.

the link's example gives "a" as the rectangle's width and "b" as its height.

your problem staes that "a" is the height and "b" is the width.

also, w = mg ... so, m = w/g.

the correct moment of inertia for the given values in your problem is

$ I = \frac{1}{3}\cdot \frac{w}{g} \cdot b^2 = \frac{wb^2}{3g}

substitute into Newton's 2nd law for rotation ...

$ F \cdot b = \frac{wb^2}{3g} \cdot \alpha

solve for \alpha ...

$ \alpha = \frac{3Fg}{wb}

substitute into the kinematics equation ...

$ \frac{\pi}{2} = \frac{1}{2} \cdot \frac{3Fg}{wb} \cdot t^2

I'll leave it to you to solve for t ... you'll get an expression for t in terms of the given values (F, w, b) and fundamental constants (g, pi).


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PostPosted: Wed, 25 Apr 2007 04:37:38 UTC 
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Thanks skeeter. Sorry i mistyped out my first answer, but still had the a and b mixed up :) Now I see how to get it.


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PostPosted: Wed, 25 Apr 2007 11:07:21 UTC 
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Sorry I have one more :(

Diagram - http://img265.imageshack.us/img265/4533 ... elpww4.jpg

Consider a bicycle wheel that initially is not rotating. A block of mass m is attached to the wheel and is allowed to fall a distance h. Assume that the wheel has a moment of inertia I about its rotation axis.

Consider the case that the string tied to the block is attached to the outside of the wheel, at a radius r_A

(a) Find omega_A, the angular speed of the wheel after the block has fallen a distance h, for this case.
Express omega_A in terms of m, g, h, r_A, and I.

(b) Now consider the case that the string tied to the block is wrapped around a smaller inside axle of the wheel of radius r_B View Figure . Find omega_B, the angular speed of the wheel after the block has fallen a distance h, for this case.
Express omega_B in terms of m, g, h, r_B, and I.

WORK DONE :

Im really not sure how to do either of these. I was going to use this formula (http://upload.wikimedia.org/math/6/c/d/ ... 84b456.png) but I couldnt apply it :( Any help would be greatly appreciated. Thanks!


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PostPosted: Wed, 25 Apr 2007 13:37:54 UTC 
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conservation of energy in both cases ...

initial PE of the mass = translational KE of the mass + rotational KE of the wheel

$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

the linear speed of the mass is related to the angular speed of the wheel by the equation v = r\omega.


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PostPosted: Wed, 25 Apr 2007 14:47:45 UTC 
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Would I be correct in saying you can rearrange the equation for w, and then sub into the formula for angular velocity?

(but I just noticed we have a v in there which is not allowed in the answer)


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PostPosted: Wed, 25 Apr 2007 21:22:18 UTC 
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bally wrote:
Would I be correct in saying you can rearrange the equation for w, and then sub into the formula for angular velocity?

(but I just noticed we have a v in there which is not allowed in the answer)


to reiterate the last statement in my previous post ...

skeeter wrote:
the linear speed of the mass is related to the angular speed of the wheel by the equation v = r\omega


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PostPosted: Thu, 26 Apr 2007 00:24:23 UTC 
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Therefore Image
can be reduced to w = (sqrt (2mgh - mg^2 - I)) as the KE of the block is 1/2 m *g^2 (due to gravity) right?

Then we can say w = (sqrt (2mgh - mg^2 - I))/r_a right?


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