65. The rotational inertia for an axis through A is I

cm

+ mh

2
A

and that for an axis through B is I

cm

+ mh

2
B

.

Using Eq. 16-29, we require

2π

I

cm

+ mh

2
A

mgh

A

= 2π

I

cm

+ mh

2
B

mgh

B

which (after canceling 2π and squaring both sides) becomes

I

cm

+ mh

2
A

mgh

A

=

I

cm

+ mh

2
B

mgh

B

.

Cross-multiplying and rearranging, we obtain

I

cm

(h

B

− h

A

) = m



h

A

h

2
B

− h

B

h

2
A



= mh

A

h

B

(h

B

− h

A

)

which simplifies to I

cm

= mh

A

h

B

. We plug this back into the first period formula above and obtain

T = 2π

mh

A

h

B

+ mh

2
A

mgh

A

= 2π

h

B

+ h

A

g

.

From the figure, we see that h

B

+ h

A

= L, and (after squaring both sides) we can solve the above

equation for the gravitational acceleration:

g =



2π

T



2

L =

4π

2

L

T

2

.