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September 30, 2014

September 30, 2014

Posted by **Jimmy** on Friday, October 31, 2008 at 2:53pm.

Question 1:

At the instant the rod is horizontal, find its angular speed. The acceleration of gravity is 9.8 m/s2 . Answer in units of rad/s.

Question 2:

At the same instant find the magnitude of its angular acceleration. Answer in units of rad/s2.

Question 3:

Still at the same instant find the magnitude of the acceleration of its center of mass. Answer in units of m/s2.

Question 4:

Finally, find the magnitude of the reaction force at the pivot (still at the same moment). Answer in units of N.

- Physics -
**Damon**, Friday, October 31, 2008 at 7:45pmI do not see the picture but assume that the rod falls with one end fixed.

The easy way to approach this is to use conservation of energy. The center of gravity of the rod has fallen 3.33 meters by the time it reaches horizontal.

It has therefore lost potential energy equal to m g (3.3) = 6.74 * 9.8 * 3.33

= 220 Joules

Therefore since the hinge is frictionless I have 220 J of kinetic energy

Call the angular velocity w, then the velocity, v, of the center of mass is 3.33 w

Ke = (1/2) m v^2 + (1/2) I w^2

I for a rod about its center is (1/12) m L^2

so

220 = (1/2)(6.74)(3.33w)^2 +(1/24)(6.74)(6.66)^2 w^2

or

220 = 49.8 w^2

w= 2.1 rad/s

(check my math!)

Now what is the angular acceleration?

The moment about the hinge is 3.33* mg

= 3.33 * 6.74 * 9.8

= 220 N m

That equals the moment of inertia about the hinge times A, the angular acceleration. I about hinge = (1/3) m L^2

so

220 = (1/3)(6.74)(6.66^2)A

so A = 2.21 rad/s^2

acceleration of middle = 3.33A = 7.35 m/s^2

force = mass * acceleration

force up = F at hinge

Force down = m g = 6.74*9.8 = 66 N

so

66 - F = 6.74*7.35

F = 16.51 N

-----------------------

check

that force * 3.33 should be (1/12) 6.74 (6.66^2) A about center

That gives A = 2.21, sure enough

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