The uniform 30 kg slender rod is being pulled by the cord th

The uniform 30- kg slender rod is being pulled by the cord that passes over the small smooth peg at point A. The moment of Mass inertia of the slender rod about the point O is:I_e = 10 (kg - m^2). If the rod has a counterclockwise angular velocity of = 8 rad/s at the instant shown Determine the tangential and normal components of reaction at the pin O Determine the angular acceleration of the rod.

Solution

STEP1:

GIVEN DATA:

Mass m= 30kg

Moment of intertia = 10

Load p= 400N

ANGULAR VELOCITY W= 8 Rad/sec

When a rigid body rotates about a fixed axis perpendicular to the plane of the body at point O, the body’s center of gravity G moves in a circular path of radius r_G. Thus, the acceleration of point G can be represented by :

a tangential component (a_G)_t = r_G a

a normal component (a_G)_n = r_G w

substitute values above equation:

r_G=0.6m

Find: The angular acceleration a and the reaction at pin O when the rod is in the horizontal position.

Plan :

Since the mass center moves in a circle of radius

1.5 m, it’s acceleration has a normal component toward O and a tangential component acting downward and perpendicular to r_G.

Apply the problem solving procedure.

Equations of motion:

+ ® åF_n = m a_n = m r_G w²
         O_n = 30(0.6)(8)² = 115.2 k N

+ ¯ åF_t = m a_t = m r_G a  
        -O_t + 30(9.81) = 30(0.6) a

+ åM_O = I_G a + m r_G a (r_G)

Þ 0.6 (15) 9.81 = I_G a + m(r_G)² a

Using I_G = (ml²)/12 and r_G = (0.6)(l), we can write:

åM_O = [(ml²/12) + (ml²/4)] a = (ml²/3) a where (ml²/3) = I_O.

After substituting: total rod length = 1m

+ 30(9.81)(0.6) = 30(1/1) a

Solving: a = 5.886 rad/s²

                O_t = 19 N

STEP2:

Since the body experiences an angular acceleration, its inertia creates a moment of magnitude, I_ga, equal to the moment of the external forces about point G. Thus, the scalar equations of motion can be stated as:

STEP3:

Consequently, we can write the three equations of motion for the body as:

å F_n = m (a_G)_n = m r_G w²

å F_t = m (a_G)_t = m r_G a

å M_G = I_G a

Note that the åM_G moment equation may be replaced by a moment summation about any arbitrary point. Summing the moment about the center of rotation O yields

åM_O = I_Ga + r_G m (a_G)_t = [I_G + m(r_G)²] a

STEP 4:

From the parallel axis theorem,
I_O = I_G + m(r_G)², therefore the term
in parentheses represents I_O.

Step 5:

Find: The angular acceleration a and the reaction at pin O when the rod is in the horizontal position.

Since the mass center moves in a circle of radius 1.5 m, it’s acceleration has a normal component toward O and a tangential component acting downward and perpendicular to r_G.
Apply the problem solving procedure.

Equations of motion:

+ åF_n = m a_n = m r_G w²          Þ O_n = 0 N

+ åF_t = m a_t = m r_G a            Þ -O_t + 30(9.81) = 30 (0.15) a

+ åM_O = I_G a + m r_G a (r_G)    Þ 0.6 (30) 9.81 = I_G a + m(r_G)² a

Step 6:

Using I_G = (ml²)/12 and r_G = (0.6m), we can write:

I_G a + m(r_G)² a = [(8×0.9²)/12 + 30(0.15)²] a = 1 a

After substituting:

22.07 = 1.35 a   Þ a = 16.4 rad/s²

-O_t + 30(9.81) = 8(0.6)a

O_t = 30(9.81) 8(0.6)16.4 = 163 N