- #1
AngelofMusic
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This was a question on one of our past exams, and our professor said it was really simple. Of course, he also said that only one person got it of the 400-some people who took the course, possibly because the scenario was very unusual.
Here's what I've done so far:
[tex]\overline{I} = 1.686 kg m^2[/tex]
When the body is at 30 degrees to the vertical:
[tex] mgcos60 * (0.57) = [\overline{I} + (375/9.81)(0.57)^2]\alpha[/tex]
Basically, I treated the body as though it were rotating about a fixed point A and found: α = 7.58 rad/s^2.
I can then find the tangential component of acceleration at mass centre G relative to A: [tex]a_{(G/A)t} = (0.57)\alpha = 4.32 m/s^2[/tex]
I figure that since the wheelchair is a rigidly connected, the acceleration at point a is equal to the acceleration of its mass center. So: [tex]a_A = 15 m/s^2[/tex] to the right.
Then I use this equation to try and find the angular velocity:
[tex]a_G = a_A + a_{(G/A)t} + a_{(G/A)n}[/tex]
[tex]a_G = 15[/tex]
The upper body of a spinal cord injured patient in the wheel chair has a mass of 375 N, a center of gravity at G, and a radius of gyration of 0.21 m. By means of the seatl belt, the body segment is assumed to be pin-connected to the seat of the wheel chair at A. If a crash causes the wheelchair to decelerate at 15 m/s^2, determine the angular velocity of the body when it was rotated to θ = 30 degrees. http://img23.photobucket.com/albums/v68/AngelOfMusic/wheelchair.jpg
Here's what I've done so far:
[tex]\overline{I} = 1.686 kg m^2[/tex]
When the body is at 30 degrees to the vertical:
[tex] mgcos60 * (0.57) = [\overline{I} + (375/9.81)(0.57)^2]\alpha[/tex]
Basically, I treated the body as though it were rotating about a fixed point A and found: α = 7.58 rad/s^2.
I can then find the tangential component of acceleration at mass centre G relative to A: [tex]a_{(G/A)t} = (0.57)\alpha = 4.32 m/s^2[/tex]
I figure that since the wheelchair is a rigidly connected, the acceleration at point a is equal to the acceleration of its mass center. So: [tex]a_A = 15 m/s^2[/tex] to the right.
Then I use this equation to try and find the angular velocity:
[tex]a_G = a_A + a_{(G/A)t} + a_{(G/A)n}[/tex]
[tex]a_G = 15[/tex]
[tex] + 4.32 [/tex][60 degrees to the vertical] + [tex](0.57)\omega [/tex][30 degrees to the vertical]
This is where I get stuck, because I don't know the direction of [tex]a_G[/tex]. Is it simply directed towards the left?
I'm not even sure if my previous steps were correct. Any help or corrections would be appreciated!
This is where I get stuck, because I don't know the direction of [tex]a_G[/tex]. Is it simply directed towards the left?
I'm not even sure if my previous steps were correct. Any help or corrections would be appreciated!
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