A long uniform rod of length 8.00 m and mass 10.00 kg is pivoted about a horizon

Question

A long uniform rod of length 8.00 m and mass 10.00 kg is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position as shown in Figure P10.61.

Figure: http://www.webassign.net/pse/p10-59.gif

(a) What is the angular speed of the rod at the instant it is horizontal?

(b) What is the magnitude of the angular acceleration of the rod at the instant it is horizontal?

(c) Find the components of the acceleration of the rod's center of mass.
ax =  
ay =  

(d) Find the components of the reaction force at the pivot.
Rx =  
Ry =

Explanation / Answer

I=M*L^2/3
a)
using energy

M*g*L/2=.5*M*L^2*^2/3
solve for

sqrt(3*g/L)=

= 1.917rad/s;

b)
=I*

M*g*L/2=M*L^2*/3
solve for
3*g/(2*L)=

= 1.837 rad/s^2;

c) -g in the y

^2*r in the x
or
3*g/2 = 14.7

d) Ry=0
Rx=M*3*g/2 = 147

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