
#1
Jun1710, 07:30 PM

P: 884

1. The problem statement, all variables and given/known data
A uniform metrerule is pivoted to rotate about a horizontal axisthrough the 40cm mark. The stick has a mass/unit length of [tex]\mu[/tex] kg/m and its rotational inertia about this pivot is [tex]0.093 \mu[/tex] kg/m^{2}. It is released from rest in a horizontal position. What is the magnitude of the angular acceleration of the rod? 2. Relevant equations An expression for the magnitude of the torque due to gravitational force about an axis through the pivot: [tex] \tau =Mg \left( \frac{L}{2} \right)[/tex] Angular accleration and torque: [tex]\sum \tau = I \alpha[/tex] 3. The attempt at a solution Since [tex]\mu = \frac{M}{L}[/tex], and L=0.4m I get [tex]\tau = 0.4 \mu g \frac{0.4}{2} = 0.78 \mu[/tex] [tex]\alpha = \frac{\tau}{I} = \frac{0.78 \mu}{0.093 \mu} = 8.43[/tex] But my answer is wrong. The correct answer must be 10.5 rad/s^{2}. Could anyone please help me? :( 



#2
Jun1710, 07:36 PM

Mentor
P: 40,875

What force provides the torque? How far does that force act from the pivot? 



#3
Jun1810, 07:04 AM

P: 884

Since the it is pivoted at the 40 cm mark, should I use 60/2 (since the mass is uniformly distributed)? But this didn't work because I ended up with [tex]\alpha = 31.6[/tex]... 



#4
Jun1810, 07:49 AM

P: 101

Rotation about a fixed axis
Typing error
torque=I*alpha torque=Fr 



#5
Jun1810, 07:50 AM

Mentor
P: 40,875





#6
Jun1810, 07:53 AM

P: 101





#7
Jun1810, 08:48 AM

Mentor
P: 40,875

(And your units are off.) 



#8
Jun1810, 05:36 PM

P: 101




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