A long uniform rod of length 7.00 m and mass 4.00 kg is pivoted about a horizont

Question

A long uniform rod of length 7.00 m and mass 4.00 kg is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position as shown in Figure P10.61.

(a) What is the angular speed of the rod at the instant it is horizontal? 2.08 rad/s

(b) What is the magnitude of the angular acceleration of the rod at the instant it is horizontal? 2.1 rad/s^2

(c) Find the components of the acceleration of the rod's center of mass.

ax=

ay=

(d) Find the components of the acceleration of the rod's center of mass.

Rx=

Ry=

Explanation / Answer

Weight of rod acts from its centre. Let denote the angle from vertical as the rod falls.

Torque = mg*L/2*Sin = 4*9.81*7/2*Sin

                                  = 137.34*Sin N-m

Inertia of rod I = mL2/3 = 4*72/3 = 65.333 kg-m2

Angular acceleration = Torque/Inertia = 137.34*Sin/65.33 = 2.1*Sin rad/s2

(a) = d/dt = 2.1*Sin

so, [d/d](d/dt) = 2.1*Sin

so, (d/d) = 2.1*Sin

Integrating on both the sides, 2/2 = -2.1*Cos + C

Since = 0 at = 0, we get, C = 2.1

So, 2/2 = -2.1*Cos + 2.1

When rod is horizontal, = /2, so 2/2 = -2.1*Cos(/2) + 2.1

So, = 2.05 rad/s

(b) = d/dt = 2.1*Sin

when rod is horizontal, = /2, so = 2.1*sin(/2) = 2.1rad/s2

(c) Ay = (L/2)* = (7/2)*2.1 = 7.35 m/s2

Ax = (L/2)*2 = (7/2)*2.052 = 14.7 m/s2

(d) Ry = m*Ay + mg = 4*7.35 + 4*9.81 = 68.64 N

Rx = m*Ax = 4*14.7 = 58.8 N

(or)

c ) ay = -g = - 9.8 m/s2

^2*r in the x
or
ax = 3*g/2
ax =14.7 m/s2

d) Ry=0
Rx = m*3*g/2

Rx = 4 * 3 * 9.8 / 2
Rx = 58.8 N

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