A rigid, massless rod has three particles with equal masses attached to it as sh

Question

A rigid, massless rod has three particles with equal masses attached to it as shown in the figure below. The rod is free to rotate in a vertical plane about a frictionless axle perpendicular to the rod through the point P and is released from rest in the horizontal position at

t = 0.

Assume that m and d are known. (Use the following as necessary: m, d, and g. Enter the magnitudes.)

(a) Find the moment of inertia of the system (rod plus particles) about the pivot.

IP =

(b) Find the torque acting on the system at t = 0.

P =   counterclockwise

(c) Find the angular acceleration of the system at t = 0.

=   counterclockwise

(d) Find the linear acceleration of the particle labeled 3 at t = 0.

a =   upward

(e) Find the maximum kinetic energy of the system.

KEmax =   

(f) Find the maximum angular speed reached by the rod.

max =   

(g) Find the maximum angular momentum of the system.

Lmax =   

(h) Find the maximum translational speed reached by the particle labeled 2.

(v2)max =   

Explanation / Answer

(a) IP = mr² = m*(d/3)² + m*(2d/3)² + m*(4d/3)² = 21md² / 9 = 7md² / 3

(b) = (m*4d/3 + m*d/3 - m*2d/3)*g = mdg

(c) = / IP = mdg / (7md²/3) = 3g / 7d

(d) a = r = (3g/7d)*2d/3 = 2g / 7

(e) Relative to a vertical position, the system has
PE = m*(4d/3 + d/3 - 2d/3)g = mdg
so that's the maximum KE (achieved when the rod is vertical).

(f) KE = ½*IP*²
mdg = ½ * (7md² / 3) * ²
g = (7d / 6) * ²
6g / 7d = ²
max = (6g / 7d)

(g) max L = IP * = (7md² / 3) * (6g / 7d) = md² * (14g / 3d)

(h) max v2 = *d/3 = (d/3) * (6g / 7d) = (2gd / 21)

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