Question

[7.] A uniform rod with mass M, length L, and moment of inertial with respect to the center of mass Icm = MLis hinged at one end (point P) so that it can rotate, without friction, around a horizontal axis. The rod is initially held at rest forming an angle with the vertical (see figure) and then released. a) Find the moment of inertia Ip of the rod with respect to point P. b) Find the magnitude of the angular velocity of the rod when it passes through the vertical. c) Find the linear speed of the center of mass when it passes through the vertical position. Write your results in terms of M, L, 0, and g. Check the units/dimensions for each answer.

Answer 1

de hem solh given that Icm = ML, @ using paralel aris theorem to calculate the moment com o og & inertia of rod about / its end Ip = Icm+md. there d=ų - If = 1 ME +M (2) ² = Ip = ML² + M1² = (1+3) ME If = 1km² => [p = 12 2 2 12 cuso A Foom adjacent diagram h = 1 4 - woso h = Ļ (1-coso) work done by Gravity W = angular velocity - Gain ink E migh = & Ip W2 ng ty (1-wso) - Mw? 39 Cl-COSO) = 2w?=> w=139 (1-CS O) angular velocity of rod when it becomes vertical 2 23 (1-asd )

soll Jd using Relation between weina Rel linear veloaty and angular velocity Mit >Vcm=_W v=rw var 30 (1-wuse) How o = √ 392 (1-6050 Now o checking unit of Ig wel * Ip = M1² unit kg-im, Mass - kg length meter (m) Dimension [Me To] # warp C1-C50) = m = 5 second w unit st es exactly sad/sec Dimension "[LT-2). IT 17 For weloaty Vcm = 4w, unit mit=m/s Dimension [L] [I'] = [LT]