- #1
NotaPhysicsMan
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Hey, here's the question:
You are an athlete on the high bar in a fully extended position 180 degrees from the right horizontal.
My weight: 667.08N
Center of mass:80.5cm
Radius of gyration: 122 cm
1)Calculate the torque at the beginning of the bar and for every 30 degrees until the rotation stops or changes direction. Assume that friction from bar produced a constant torque a 30 Nm.
http://www.geocities.com/mvxraven/gymnastdiagram.JPG"
So from my diagram:
sum(torque) at horiziontal=0
0=(F1xd) + (-(30Nm))
=(667.05*0.805)-30
=506.9 Nm
I can tell it's angular velocity is increasing by: angular acceleration=T/I
in this case a=+ve.
-----------------------------------
sum(torque) at 30degrees=0
0=(F1xd) + (-(30Nm))
=(667.05)(0.805cos30)-30
=435.03 Nm
a=+ve, so still not slowing down or changing direction
-----------------------------------
sum(torque) at 60 degrees=0
0=(F1xd) + (-(30Nm))
=(667.05)(0.805cos60)-30
=238.49 Nm
a=+ve still.
-----------------------------------
sum(torque) at 90 degrees=0
0=(F1xd) + (-(30Nm))
=(667.05)(0)-30
=-30Nm
Angular acc (a)
a=-ve, so it means a change in direction or stopping?
2)calculate the angular velocity at each position assuming that the torque from the previous position was applied for a period of 0.1s.
Ok, so this is just w=@2-@1/t
so should I be taking the @1 as 0, or 180 degrees?
if @1=0 then @2=30 degrees or 0.523rads.
And I just 0.523-0/0.1s = 5.23rads/s
--------------------------------------
and the next one is similar @2=60, and @1=0
so 1.04rads/0.1s or should it be 1.04/0.2s?
...etc
You are an athlete on the high bar in a fully extended position 180 degrees from the right horizontal.
My weight: 667.08N
Center of mass:80.5cm
Radius of gyration: 122 cm
1)Calculate the torque at the beginning of the bar and for every 30 degrees until the rotation stops or changes direction. Assume that friction from bar produced a constant torque a 30 Nm.
http://www.geocities.com/mvxraven/gymnastdiagram.JPG"
So from my diagram:
sum(torque) at horiziontal=0
0=(F1xd) + (-(30Nm))
=(667.05*0.805)-30
=506.9 Nm
I can tell it's angular velocity is increasing by: angular acceleration=T/I
in this case a=+ve.
-----------------------------------
sum(torque) at 30degrees=0
0=(F1xd) + (-(30Nm))
=(667.05)(0.805cos30)-30
=435.03 Nm
a=+ve, so still not slowing down or changing direction
-----------------------------------
sum(torque) at 60 degrees=0
0=(F1xd) + (-(30Nm))
=(667.05)(0.805cos60)-30
=238.49 Nm
a=+ve still.
-----------------------------------
sum(torque) at 90 degrees=0
0=(F1xd) + (-(30Nm))
=(667.05)(0)-30
=-30Nm
Angular acc (a)
a=-ve, so it means a change in direction or stopping?
2)calculate the angular velocity at each position assuming that the torque from the previous position was applied for a period of 0.1s.
Ok, so this is just w=@2-@1/t
so should I be taking the @1 as 0, or 180 degrees?
if @1=0 then @2=30 degrees or 0.523rads.
And I just 0.523-0/0.1s = 5.23rads/s
--------------------------------------
and the next one is similar @2=60, and @1=0
so 1.04rads/0.1s or should it be 1.04/0.2s?
...etc
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