Physics problem --- finding acceleration and linear velocity of a rod given mass and length?

Question

A 0.22 kg rod of length 65 cm is suspended by a frictionless pivot at one end. It is held horizontal and released.

(a) Immediately after it is released, what is the acceleration of the center of the rod?

____m/s2

(b) Find the initial acceleration of a point on the end of the rod.

____m/s2

(c) What is the linear velocity of the center of mass of the rod when it is vertical?

____m/s

Answer 1

Let:

m be the mass of the rod,

L be its length,

alpha be the angular acceleration,

w be the angular velocity,

g be the acceleration due to gravity.

I be the moment of inertia of the rod about the pivot.

mgL / 2 = I alpha

I = mL^2 / 3

Eliminating I:

alpha = 3g / 2L

(a)

The acceleration of the centre is:

alpha L / 2 = 3g / 4

= 7.36 m/s^2.

(b)

The acceleration of the end is:

alpha L = 3g / 2

= 14.7 m/s^2.

(c)

Equating initial PE to final KE:

I w^2 / 2 = mg L / 2

w^2  = 3g / L

The linear velocity of the centre of mass is:

Lw / 2 = L sqrt(3g / L) / 2

sqrt(3gL) / 2

= 2.19 m/s.