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

A long uniform rod of length 6.66 m and mass 6.74 kg is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position as in the figure.

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 7:45pm
I 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

## Answer this Question

## Related Questions

- Physics - A thin uniform rod (length = 1.3 m, mass = 3.2 kg) is pivoted about a ...
- Physics (please help!!!!) - A thin uniform rod (length = 1.3 m, mass = 4.1 kg) ...
- Physics - A nonuniform 2.0-kg rod is 2.0 m long. The rod is mounted to rotate ...
- physics - A long uniform rod of length 4.76 m and mass 5.48 kg is pivoted about ...
- physics - A long uniform rod of length 4.76 m and mass 5.48 kg is pivoted about ...
- Physics - A mass of 100g is attached to one end of a massless rod, 10cm in ...
- physics - A thin non-uniform rod of length L=2.00 m and mass M=9.00 kg is free ...
- physics - A uniform thin rod of length 0.55 m and mass 5.5 kg can rotate in a ...
- AP Physics - A uniform thin rod of length 0.3m and mass 3.5kg can rotate in a ...
- physics - A uniform rod of mass M = 0.245kg and length L = 0.49m stands ...

More Related Questions