Linear acceleration of a rod fixed to a hinge at one end

[mozartkart]

Homework Statement

a uniform stick with a mass of 1.00 kg and length of 1.00m has a fixed hinge at one end. It is released at an angel of 60 degrees above the horizontal and it drops with normal gravity. What is the magnitude of the linear acceleration of the point at the free end just after release? express the result as a percentage of g (gravity)

The Attempt at a Solution

attempt

T=r*F*sin(60) T=Inertia*alpha =(mr²)*(a/r)=mra
mra=r*m*g*sin(60)
a=g*sin(60)

Is this a correct method to approach this question with or do I have to use the inertia equation for a rod rotating about one end which is 1/3 *m*L² ? If that's the case then would this be right?

(1/3*m*L²)*a/r=rmgsin(60)
a=(r²*g*3*sin(60)) / L

 

[tiny-tim]

welcome to pf!

hi mozartkart! welcome to pf!

… do I have to use the inertia equation for a rod rotating about one end which is 1/3 *m*L² ? If that's the case then would this be right?

(1/3*m*L²)*a/r=rmgsin(60) …

basically, yes

(except why is there an L and an r? and don't forget the weight is in the middle )

 

[mozartkart]

there's a L and a r because the equation for alpha is a/r and the inertia equation is 1/3 m L². Or am I approaching that wrong?ps thank you for helping me :)

 

[tiny-tim]

there's a L and a r because the equation for alpha is a/r and the inertia equation is 1/3 m L².

ah, but then you need to choose either L or r, and stick to it when you write your equation!


(also, i think you're using two different r's )

 

[mozartkart]

ok :) I figured it out. Thank you so much for your help