A long uniform rod of length L and mass M is pivoted about a frictionless, horiz
ID: 1485726 • Letter: A
Question
A long uniform rod of length L and mass M is pivoted about a frictionless, horizontal pin through one end. The rod is nudged from rest in a vertical position as shown in the figure below.
(a) At the instant the rod is horizontal, find its angular speed. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)
? =
(b) At the instant the rod is horizontal, find the magnitude of its angular acceleration. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)
? =
(c) At the instant the rod is horizontal, find the x and y components of the acceleration of its center of mass. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)
ax =
ay =
(d) At the instant the rod is horizontal, find the components of the reaction force at the pivot. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)
Rx =
Ry =
Explanation / Answer
a) Apply conservation of energy
final kinetic energy = initial potentail energy
(1/2)*I*w^2 = m*g*h
(1/2)*(m*L^2/3)*w^2 = m*g*(L/2)
m*L^2*w^2/3 = m*g*L
L*w^2/3 = g
w = sqrt(3*g/L) <<<<<<<-----------Answer
b) Apply, net torque Tnet = I*alfa
alfa = Tnet/I
= m*g*(L/2)/(m*L^2/3)
= (3/2)*g/L <<<<<<<-----------Answer
c) ax = r*w^2
= (L/2)*w^2
= (L/2)*(3*g/L)
= (3/2)*g <<<<<<<-----------Answer
ay = r*alfa
= (L/2)*(3/2)*(g/L)
= (3/4)*g <<<<<<<-----------Answer
d) Rx = m*ax
= (3/2)*m*g <<<<<<<-----------Answer
Ry = m*g - m*ay
= m*g - (3/4)*m*g
= (1/4)*m*g <<<<<<<-----------Answer