Calculating Forces and Acceleration in a Castle Bridge Mishap

[pegasus24]

Homework Statement

Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed in his quest to improve communication between damsels and dragons. Unfortunately his squire lowered the draw bridge too far and finally stopped it 20.0° below the horizontal. Lost-a-Lot and his horse stop when their combined center of mass is 1.00 m from the end of the bridge. The uniform bridge is 8.00 m long and has a mass of 1700 kg. The lift cable is attached to the bridge 5.00 m from the hinge at the castle end, and to a point on the castle wall 12.0 m above the bridge. Lost-a-Lot's mass combined with his armor and steed is 1000 kg. Suddenly, the lift cable breaks! The hinge between the castle wall and the bridge is frictionless, and the bridge swings freely until it is vertical.
(a) Find the angular acceleration of the bridge once it starts to move.(rad/s2)
(b) Find the angular speed of the bridge when it strikes the vertical castle wall below the hinge.(rad/s)
(c) Find the force exerted by the hinge on the bridge immediately after the cable breaks.
(d) Find the force exerted by the hinge on the bridge immediately before it strikes the castle wall.

Homework Equations

a) $$\tau=I\alpha$$
b) $$\alpha=\frac{d\omega}{dt}$$
c) T = 0
d) $$\tau$$ = r x F

The Attempt at a Solution

I think these are the equations i should use in solving this problem. But I have no clue where to start.
I think $$\tau$$ can be found using (rFsin$$\theta$$) but what about inertia? Could you please giv me a hint to get started on this.. Thank you.

 

[tiny-tim]

hi pegasus24!

you'll also need to use F = ma

find the moment of inertia by adding the moment of inertia of the bridge to the moment of inertia of Sir Lost-a-Lot

 

[pegasus24]

Moment of Inertia of bridge:
Ib = mr^2
but how to calculate moment of inertia for lost-a-lot?

 

[tiny-tim]

hi pegasus24!

(try using the X² and X₂ icons just above the Reply box )

the moment of inertia of the bridge certainly isn't mr²

you need to learn most of the moments of inertia at http://en.wikipedia.org/wiki/List_of_moments_of_inertia"

 

[1994Bhaskar]

Hi tiny-tim.We don't know the structure of bridge-whether it is like a thin rod or it is cylindrical?
How come we find it's moment of inertia then?

 

[tiny-tim]

hi pegasus24!

always read the question carefully …

… The uniform bridge …

 

[pegasus24]

hi pegasus24!

(try using the X² and X₂ icons just above the Reply box )

the moment of inertia of the bridge certainly isn't mr²

you need to learn most of the moments of inertia at http://en.wikipedia.org/wiki/List_of_moments_of_inertia"

Yes I went through this link. So the
I_b = (m/3)L²
Is this right? Then what about inertia for the person and his horse? How to calculate that? The only way I know is I = (1000g)(L-1)².

 

[tiny-tim]

I_b = (m/3)L²

yup!

and in questions like this, Sir and his horse can be treated as a point particle, so the moment of inertia is just mr²

 

[pegasus24]

When i calculate torque, should I take into account the tension in the cable?

 

[tiny-tim]

what cable?

it's broken! ​

 

[pegasus24]

[tex]\tau = (1700kg)(9.8m/s²)(5m) + (1000kg)(9.8m/s²)(7m) = 151900kgm²/s²[/tex]
[tex]I = I_b + I_g = \frac{1}{3}(1700kg)(8m)² + (1000kg)(7)² = 85266.67kgm²[/tex]
[tex]\alpha = \frac{\tau}{I} = 1.78rad/s²[/tex]

But this $$\alpha$$ is wrong.. where did i go wrong?

 

[tiny-tim]

hi pegasus24!

(you can't use SUP and SUB in tex, you must use ^ and _ )

\tau = (1700kg)(9.8m/s²)(5m) + (1000kg)(9.8m/s²)(7m) = 151900kgm²/s²
I = I_b + I_g = \frac{1}{3}(1700kg)(8m)² + (1000kg)(7)² = 85266.67kgm²
\alpha = \frac{\tau}{I} = 1.78rad/s²

erm … shouldn't that (5m) be (4m) ?

 

[pegasus24]

I tried with 4m but it is still wrong..

 

[tiny-tim]

oh! you didn't use the 20°

 

[pegasus24]

I have already tried that but the answer is too small.. It should be 1. something but when i use this 20 degree angle i get 0.07 something... which is totally wrong..