A long uniform rod of length L and mass M ispivoted about a horizontal, friction
ID: 1722363 • Letter: A
Question
A long uniform rod of length L and mass M ispivoted about a horizontal, frictionless pin through one end. Therod is released from rest in a vertical position. (a) At the instant the rod is horizontal, find its angularspeed. (Use L for L, M for M, and g for gravity,as necessary.)= 1rad/s
(b) Find the magnitude of its angular acceleration.
= 2rad/s2
(c) Find the x and y components of theacceleration of its center of mass.
ax = 3m/s2
ay = 4m/s2
(d) Find the components of the reaction force at the pivot.
Rx = 5N
Ry = 6N A long uniform rod of length L and mass M ispivoted about a horizontal, frictionless pin through one end. Therod is released from rest in a vertical position. (a) At the instant the rod is horizontal, find its angularspeed. (Use L for L, M for M, and g for gravity,as necessary.)
= 1rad/s
(b) Find the magnitude of its angular acceleration.
= 2rad/s2
(c) Find the x and y components of theacceleration of its center of mass.
ax = 3m/s2
ay = 4m/s2
(d) Find the components of the reaction force at the pivot.
Rx = 5N
Ry = 6N
Explanation / Answer
let us consider that the system is the rod andthe earth (a) as only conservative forces act within thesystem of the rod and the earth we get that E = 0 KEf + PEf =KEi + PEi (1 / 2) I 2 + 0 = 0 + M g(L / 2) I = (1 / 3) m L2 so the angular velocity will be = (3 g / L) (b) as = I M g (L / 2) = (M L2 / 3) the angular acceleration will be = 3 g / 2 L (c) the acceleration in the x-direction is givenby ax = ar = - r2 = - (L / 2)2 = - 3 g / 2 ay = - at = - r = - (L /2) = - 3 g / 4 (d) according to the newtons second law ofmotion we get that Rx = M ax = - 3 M g /2 Ry - M g = M ay =- 3 M g / 4 Ry = M g / 4 so the angular velocity will be = (3 g / L) (b) as = I M g (L / 2) = (M L2 / 3) the angular acceleration will be = 3 g / 2 L (c) the acceleration in the x-direction is givenby ax = ar = - r2 = - (L / 2)2 = - 3 g / 2 ay = - at = - r = - (L /2) = - 3 g / 4 (d) according to the newtons second law ofmotion we get that Rx = M ax = - 3 M g /2 Ry - M g = M ay =- 3 M g / 4 Ry = M g / 4