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9.3 Center of Gravity

W1 x1 + W2 x2 + L xcg = W1 + W2 + L DEFINITION OF CENTER OF GRAVITY The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated.

9.3 Center of Gravity

Finding the center of gravity of an irregular shape.

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

Consider a powered model airplane on a massless string.

FT = maT ! = FT r

aT = r! m

" = (mr )! 2

Moment of Inertia, I

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

" 1 = (m1r12 )!

#" = # (mr )! 2

Net external torque

Moment of inertia

" 2 = (m r )! 2 2 2

M

" N = (mN rN2 )!

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

ROTATIONAL ANALOG OF NEWTON’S SECOND LAW FOR A RIGID BODY ROTATING ABOUT A FIXED AXIS

& Moment of Net external torque = $$ % inertia

# & Angular # !! ' $$ !! " % acceleration "

#" = I ! Requirement: Angular acceleration must be expressed in radians/s2.

( )

I = ! mr 2

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

Example 9 The Moment of Inertial Depends on Where the Axis Is. Two particles each have mass and are fixed at the ends of a thin rigid rod. The length of the rod is L. Find the moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at (a) one end and (b) the center.

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

(a)

( )

2

2

I = ! mr 2 = m1r12 + m2 r22 = m(0 ) + m(L )

m1 = m2 = m

2

I = mL

r1 = 0 r2 = L

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

( )

(b) I = ! mr 2 = m1r12 + m2 r22 = m(L 2 )2 + m(L 2 )2

m1 = m2 = m

2

I = mL 1 2

r1 = L 2 r2 = L 2

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

I = (2/5)MR2

I = MR2

I = (1/2)MR2

I = (1/12)ML2

I = (1/3)ML2

I = (7/5)MR2

I = (2/3)MR2

I = (1/12)ML2

I = (1/3)ML2

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

Example 12 Hoisting a Crate A motor is used to lift a crate with the dual pulley system shown below. The combined moment of inertia of the dual pulley is 46.0 kg·m2. The crate has a mass of 451 kg. A tension of 2150 N is maintained in the cable attached to the motor. Find the angular acceleration of the dual pulley and the tension in the cable connected to the crate.

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis equal

" F = T # y 2 ! mg = ma y a y = l 2!

$" = T l

T2 = mg + ma y

1 1

# T2 l 2 = I!

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

T1l 1 " (mg + ma y )l 2 = I! a y = l 2!

T1l 1 " (mg + ml 2! )l 2 = I! T1l 1 " mgl 2 #= I + ml 22

( 2150 N )(0.600 m )" (451 kg )(9.80 m s 2 )(0.200 m ) = = 6.3 rad 2 2 46.0 kg ! m + (451 kg )(0.200 m )

s2

9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis

Tension in the cable connected to the crate:

T2 = mg + may = mg + ml2α = (451 kg)(9.80 m/s2) + (451 kg)(0.200 m)(6.3 rad/s2) = 5000 N

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