Solving for Angular Acceleration & Speed of Rod in Freefall

[Juniper7]

Homework Statement

Consider a uniform rod of mass 12kg and length 1.0m. At its end, the rod is attached to a fixed, friction-free pivot. Initially the rod is balanced vertically above the pivot and begins to fall (from rest) as shown in the diagram. Determine,
a) the angular acceleration of the rod as it passes through the horizontal at B.
b) the angular speed of the rod as it passes through the vertical at C.

Homework Equations

PE = mgh
KE = 1/2Iω²
I_rod = 1/3ml²
ω² = ω_o² + 2αθ

The Attempt at a Solution

a) PE = KE
mgh = 1/2Iω²
(12kg)(9.81m/s²)(1.0m) = (0.5)(1/3)(12kg)(1.0m)ω²
ω = 7.668rad/s
Can I use the length of the rod as its height?

ω² = ω_o² + 2αθ
(7.668)² = 2α90° (90° = 1.571rad)
α = 18.7rad/s²
is that right?

b) ω² = ω_o² + 2αθ
ω² = (7.668rad/s)² + 2(18.7rad/s²)(1.571rad)
assumptions: the α will stay the same and I'm trying to calculate from point B to C so θ = 90°
ω = 10.84rad/s

Thanks in advance!

 

[tms]

Use the center of mass of the rod.

 

[Tanya Sharma]

a) PE = KE
mgh = 1/2Iω²
(12kg)(9.81m/s²)(1.0m) = (0.5)(1/3)(12kg)(1.0m)ω²
ω = 7.668rad/s
Can I use the length of the rod as its height?

ω² = ω_o² + 2αθ
(7.668)² = 2α90° (90° = 1.571rad)
α = 18.7rad/s²
is that right?

b) ω² = ω_o² + 2αθ
ω² = (7.668rad/s)² + 2(18.7rad/s²)(1.571rad)
assumptions: the α will stay the same and I'm trying to calculate from point B to C so θ = 90°
ω = 10.84rad/s

This is incorrect .

The relation in red is applicable when the angular acceleration is constant which is not the case in this problem.

Write torque equation for the rod when it is horizontal.

 

[Juniper7]

This is incorrect .

The relation in red is applicable when the angular acceleration is constant which is not the case in this problem.

Write torque equation for the rod when it is horizontal.

Ok, so:

τ = Iα
τ = 1/3ml²α

τ = rF
F = mg
τ = rmg
rmg = 1/3ml²α
(1.0m)(12kg)(9.81m/s²) = (1/3)(12kg)(1m)²α
α = 29.4rad/s²
This seems rather high...

would the radius of the rod be its length because it is pivoting on its end? I think i did something wrong... thanks for the help

 

[Tanya Sharma]

would the radius of the rod be its length because it is pivoting on its end?

1) What is the point of application of the force mg which is producing the torque about the pivot ?
2) What is the distance between this point of application and the pivot ?

 

[Juniper7]

1) What is the point of application of the force mg which is producing the torque about the pivot ?
2) What is the distance between this point of application and the pivot ?

1) Would that be at the end of the rod, the opposite end of the pivot?
2) if the point of application is the end, then it would be 1.0 m and the radius should be 1.0m, right?

 

[Tanya Sharma]

1) Would that be at the end of the rod, the opposite end of the pivot?
2) if the point of application is the end, then it would be 1.0 m and the radius should be 1.0m, right?

No.

Where does weight of a body act ?

 

[Juniper7]

No.

Where does weight of a body act ?

Oh, the centre of mass? so the radius would be half of the length, 0.5m, right?

 

[Tanya Sharma]

Oh, the centre of mass? so the radius would be half of the length, 0.5m, right?

Correct.

 

[Juniper7]

Thank you!