Simple pendulum: Determine the velocity

[dirk_mec1]

Homework Statement

A homogeneous bar with length 0.6 m and mass m = 2 kg is fixed to a wall via a hinged connection in the vertical plane. At the end of the bar a constant force F acts of 150 N. The bar is released from the vertical equibrilium position. Determine the velocity (of the COG) and the angular velocity of the bar when the bar reaches a horizontal postition.

The answer from the book is: [itex]\omega [/itex] = 26.4 rad/s and [itex]v_{COG}[/itex] = 7.9 m/s

Homework Equations

Work moment : [tex] M = I \theta [/tex]
Rotational energy:

[tex] \frac{1}{2} I \omega ^2 [/tex]

The Attempt at a Solution

[tex] M \theta =\frac{1}{2} I \omega ^2 =\frac{1}{2} (\frac{1}{3}ML^2) \omega ^2 [/tex]

[tex] 150 \cdot0.6 \cdot \pi/2= 1/2 \cdot 1/3\cdot 2 \cdot 0.6^2 \cdot \omega^2 \rightarrow \omega =\ 34.32 rad/s [/tex]

What mistake am I making?

 

[DEvens]

I see no gravity here.

 

[andrevdh]

Potential energy

 

[dirk_mec1]

The gravity is not mentioned in the exercise but let's add it:

[tex] M \theta =\frac{1}{2} I \omega ^2 + mg\Delta h \rightarrow\ \omega = 23.5\ rad/s [/tex]

Why is my answer not agreeing with the one from the book?

 

[haruspex]

The question is not clear. Which way is F acting initially, and which way does it act later?
It says the force is constant, not that its magnitude is constant. That suggests its direction is constant.

 

[dirk_mec1]

Yeah you're right the question is posed wrongly.. I found out I get the right answer if I assume the force acts horizontally.