A stick of uniform density with mass M = 7.8 kg and length L = 0.8 m is pivoted

Question

A stick of uniform density with mass M = 7.8 kg and length L = 0.8 m is pivoted about an axle which is perpendicular to its length and located 0.11 m from one end. Ignore any friction between the stick and the axle.

a) What is the moment of inertia of the stick about this axle?

Iaxle = ?? kg m2

The stick is held horizontal and then released.

b) What is its angular speed as it passes through the vertical

w = ?? rad/s

c) What is its angular acceleration as it passes through the vertical position?

a = ?? rad/s2

d) What is the magnitude of the vertical component of the force exerted by the stick on the axle when the stick passes through the vertical?

|Fvertical| = ?? N

Explanation / Answer

Rotational dynamics are analogous to translational dynamics. Moment of inertia is like mass. Take an object and break it into simple components that the moment is easy to calculate. Then sum the moments of the parts to get the moment of the whole. This object can be modeled as two rods rotating about their end. One is 0.23 m long and has mass 7.8*0.11/0.8 = 1.0725 The other is 0.8-0.11 m long and has mass 7.8*(0.8-0.11)/0.8 For a rod pivoting about one end I=m*L^2/3 So take each piece independently first the short one 7.8*0.11*0.11^2/(3*0.8) 0.0043 and the longer one 7.8*(0.8-0.11)^3/(3*0.8) 1.0676 add the two to get the moment for the object a) 1.07195 b) I will use energy for this Each segment of the rod will either gain or lose PE by the change in elevation of it's center of mass. I will look at each segment independently g*7.8*((0.8-0.11)^2-0.11^2)/(2*0.8) 22.1676 J of PE lost That gets converted to angular momentum of .5*I*?^2 equate and solve for ? ?=sqrt(2*22.1676/1.0676) 6.444 rad/s c) This is asked to see if you understand that St=I*a When the rod is vertical, St=0, therefore, a=0 d) The vertical force will have two components: The weight m*g, downward and the centripetal force. Again, I will look at each segment independently and use F=m*?^2*r we have ?=6.444, which is the same fore either segment for each segment, the mass we calculated above, and the r is the center of mass for that segment small 6.444^2*7.8*0.11^2/(2*0.8) = 2.4494 This is upward large 6.444^2*7.8*(0.8-0.11)^2/(2*0.8) = 96.37 this is downward take positive as downward 2.4494 N add the weight 96.37 N for a total force downward of 98.81N

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