- #1

- 146

- 0

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

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 [email protected]@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 [email protected]@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

Last edited by a moderator: