Question

In the figure, a thin uniform rod (mass 4.6 kg, length 5.0 m) rotates freely about a horizontal axis A that is perpendicular to the rod and passes through a point at a distance d = 1.4 m from the end of the rod. The kinetic energy of the rod as it passes through the vertical position is 18 J. (a) what is the rotational inertia of the rod about axis A? (b) what is the (linear) speed of the end B of the rod as the rod passes through the vertical position? (c) At what angle θ will the rod momentarily stop in its upward swing?

Answer 1

given

m = 4.6 kg ,

length L = 5 m

d = 1.4 m

KE = 18 J

a )

using equation to find the inertia is " I "

I = ( m L²/12 ) + m ( L/2 - d )²

I = ( 4.6 X 5²/12 ) + 4.6 X ( 5/2 - 1.4 )²

I = 9.583 + 5.566

I = 15.149 kg m²  

b )

KE = I $\omega$ ² / 2

then $\omega$ = ( 2 KE / I )^1/2

$\omega$ = ( 2 X 18 / 15.149 )^1/2

$\omega$ = 1.541 rad/sec

and the speed is " V "

V = $\omega$ ( L - d )

V = 1.541 X ( 5 - 1.4 )

V = 5.5476 m/sec

c )

KE = m g ( L/2 - d ) ( 1 - cos $\theta$ )

18 = 4.6 X 9.8 X ( 5/2 - 1.4 ) ( 1 - cos $\theta$ )

18 = 49.588 X ( 1 - cos $\theta$ )

( 1 - cos $\theta$ ) = 0.3629

cos $\theta$ = 0.6371

$\theta$ = 50.424^o