netP I P the rotational second law form for a rotation of the rod around its

# Netp i p the rotational second law form for a

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net,P = I P " , the rotational second law form for a rotation of the rod around its end pivot point P. The first ingredient is I P = 1 3 mL 2 for the rod around the axis perpendicular to and through its end (which we already used in Ch. 9++ to calculate the rotational kinetic energy). The torque due to the rod’s weight is + mg L 2 cos ! because 1) we now use the convention “CW is positive,” 2) the CM is at L/2, and 3) F ! = mgcos " (or, if you prefer, you can use the equivalent torque formula, FRsin( ! 2 + " ) = mg L 2 cos " , which gives the same answer). The rotational second law, ! net,P = I P " , now reads + mg L 2 cos ! = + 1 3 mL 2 " or, simplifying and rearranging, ! = 3 2 g L cos " C Inserting this into F P, tangential + mgcos ! = + m " L / 2 , we have F P, tangential = ! mgcos " + m 3 2 ( g L cos " ) L / 2 , or F P, tangential = ! 1 4 mgcos " D (example on next page)
Mechanics Cycle 3 Chapter 12++ 12-20 P radial is horizontal now tangential is vertical now Example: Suppose a thin uniform rod of mass m and length L has its frictionless hinge pivoted on a horizontal axis. It is lifted up to the angle of 30 o with respect to the horizontal plane and is released from rest, rotating downward due to gravity. As a result, it swings down through # = 0 having some nonzero angular velocity at that horizontal position. Notice from our general formula C obtained from the rotational second law that ! 0 = 3 3 4 g L at the initial 30 o position and ! = 3 2 g L at the horizontal position. What are the forces due to the pivot on the rod at the horizontal position ( # =0) , after falling from rest at the initial angle of 30 0 ? 1) First, let’s go after F P, radial where we need ! 2 . Note from the figure on the right, this is the horizontal force on the rod due to the pivot. With ! 0 = 0 at # 0 = 30 o , the energy result B , ! 2 = ! 0 2 + 3 g L (sin " 0 # sin " ) , reduces to: ! 2 = " 3 g L (0 " sin 30 o ) = + 3 2 g L So formula A, F P, radial - mgsin ! = " m # 2 L / 2 turns into F P, radial ! 0 = ! m 3 2 g L L / 2 , or F P, radial = ! 3 4 mg at " = 0 (this is only true for the fall from rest at 30 0 ) Indeed, the pivot has to pull back to the left on the rod as it swings down (to make it go in a circle!) This component, - ! mg, is horizontal and points to the left , like a good centripetal force should for the rod in a horizontal position. (continuing next to the tangential force component) ! 0 = 0, " 0 ! , "
Mechanics Cycle 3 Chapter 12++ 12-21 2) Now for F P, tangential , which, at # = 0, is a vertical component (see the previous figure). Plugging into formula D, F P, tangential = ! 1 4 mgcos " , we get F P, tangential = ! 1 4 mg at " = 0 (but this is independent of the initial conditions) From the minus sign (down was positive), the pivot thus pushes vertically up on the rod in reaction to its weight, at # = 0. But notice it is NOT ! 1 2 mg as it would be if the rod were supported on both ends and wasn’t rotating down. The interesting fact is that the pivot supports less than half the weight when the rod is swinging down!!

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