Question

A long, thin rod of mass M and length L is standing straight up on a table. Its lower end rotates on a frictionless pivot. A very slight push causes the rod to fall over.

A. As it hits the table, what is the angular velocity of the tip of the rod?

B. What is the speed of the tip of the rod?

Answer 1

here,

applying conservation of rotational energy to this situation

at the start, there is only PE and no KE

just before impact, all the PE has converted to KE

at the start, the center of mass is half way up, so the PE of this system is 1/2 M g L where L is the length

the KE at the end is 1/2 I w^2 where I is the moment of inertia and w is the angular velocity

I for a stick pivoted at an end = 1/3 m L^2, so we have

1/2 *m *g *L = 1/2 * 1/3 *m *L^2 w^2

w = Sqrt[3 g/L]

A)

w = sqrt( 3*9.8 /L)

w = 5.42/sqrt(L)

the angular velocity of tip ia 5.42/sqrt(L) rad/s

B)

the speed of the Tip , v = l * w

v = L * 5.42/sqrt(L)

v = 5.42 * sqrt(L)

the speed of the tip of rod is 5.42 * sqrt(L) m/s

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