A metal rod of length 0.66 m is hanging vertically from one end and is free to p
ID: 1628615 • Letter: A
Question
A metal rod of length 0.66 m is hanging vertically from one end and is free to pivot around that end. The center of mass of the rod (cm) is at its geometrical center. The hanging end of the rod is struck and given a linear speed of v = 2.8 m/s. The rod rotates upward around its pivot and comes to a stop at some angle relative to the vertical. The center of mass (cm) rises some distance h during this. (a) What is h? (b) What would the minimum initial linear speed of the end of the rod have to be in order to make the rod go all the way around?Explanation / Answer
length of rod(L)=0.66m
velocity given to the rod(v)=2.8m/s
let the center of mass of the rod raises by h
applying the law of conservation of energy we get
(1/2)xmv^2=mgh
where h=(L/2)(1-costheta)
on solving we get costheta=1-(v^2/gL)
h=1-costheta=1-(1-(v^2/gL))
h=v^2/(gL)
=(2.8^2)/(9.8x0.66)
=1.21m
b)to complete the loop the center of mass of rod should reach the hightest point
(1/2)mv^2=mgL
v=sqrt(2gl)
=(2x9.8x0.66)^(1/2)
=3.6m/s
therefore should be initial speed of 3.6m/s to complete the loop