A long uniform rod of length 10.00 m and mass 8.00 kg is pivoted about a horizon

Question

A long uniform rod of length 10.00 m and mass 8.00 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? (b) What is the magnitude of the angular acceleration of the rod at the instant it is horizontal? (c) Find the components of the acceleration of the rod's center of mass. (d) Find the components of the reaction force at the pivot.

Explanation / Answer

a) using energy

Mg(L/2) = (1/2)(ML2/3)2

Solve for

= (3g/L)

= (3*9.8/10)

= 1.715 rad/s

b)

b) Mg(L/2) = (ML2/3)

Solve for

= 3g/(2L)

= 3*9.8/(2*10)

= 1.47 rad/s2

c)

Acceleration is

ax = r ^2

    = (L/2) (3g/L)

    = 3g /2

   = 3*9.8 /2

   = 14.7 m/s2

ay = -(L/2)

    = -3g/4

   = -3*9.8 / 4

   = - 7.35 m/s2

d) Newton second law

Rx = M ax = M(3g/2)

     = - 3Mg/2
     = - 3*8*9.8/2

    = -117.6 N
Ry - Mg = M ay

Ry = Mg + May

    = Mg + M (-3g/4)

    = Mg /4

    = 8*9.8 /4

    = 19.6 N

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