Question

The pendulum shown below consists of a 100-kg sphere (r= 0.2 m) attached to a 20-kg slender rod OB (L = 2.0 m). It swings about a pin at 0. At the instant 0 = 60°, the pendulum has an angular velocity of oj = 2 rad/s as shown. Determine the angular velocity of the pendulum when 0 = 0° (pendulum is vertical). olso L = 2 m 0 = 60° 20 kg 20 kg 70.4 m 2.2 m -100 kg

Answer 1

Here we are assuming that intil position of rod is when angle with rod equal to 60° snd final position of of rod is when angle eqiual to 0°.Then calculate total energy of system at intial and final position, we will conserve these energies.
Here we will first calculate the gravitational potential energy of the system, onen o= 60° As the mass of body is concentrated at it's center of mass. a center of mass of rod is at it's midpoint and center of dop mass of Sphere is at it's center the hao consider as shown reference level of the figure in

h+ 1 ts neight Los height b19_- Iler T - 511---- Let Q be the center of mass of rod .: DO APTQ VII Coso PT coso & PT 212 . PT ll coso PT + Iva COS 60°

OF to let hi be the height of center mass OF rod - H4 -- 0 because height is in down coorod direction from ho center the of now mass is at Similasly point R . In APSR cas 60° se na ps 7 (4+0) 105600 ps (1+0) Y2

center of Let ha be the height of mass of sphere i ha 7 - ( 1+0) --- 2 as been is is in doconcoqod height direction for point R Intial groavitation potential eneogy of System Ų > magh, + Mg ghz here my is mass of rod ms s is mass of sphere Us Mod (, + msg f(x 0 ) -- 2 (Fomos 0) ya - [met, meg turing U; + - [Chopin a meg cury) --- m

Now here rod will only rotate about point of suspension It will not have translational kinetic energy It will have only rotational Kinetic ୧୦୫୨ As rod is released from rest wielo Intial Rotational KE OF System is zero R.KE, a 12 I W; 2 1 1 202 ao . (RKE): 70 -- ( Final Now we potential will calculate energy of -- -- hyo a gravitation system in coken of oo final condition 17 & for & heigney of center of mass of roda in this case is le again a negative because height is. 0 iD do conward direction

Now the sphere R in height this of case center of mass is - (l+8) of " Final gravitational potential energy of system Up & mag hot mo ghs Uf Mag (Wa) + (mag - (Stry] Up + - [mage + mos g (4+0] -- ☺ Now to calculate the rotatational Kinetic energy we first need to calculate momenter inertia of given system As the both system rotate about point of suspension; - objects we coin apply parallel anis For rod theorem

THEP at As the axis of rotation passiong through the center and axis of rotation passing through the point of supension are Parallel to each other I let IP be the moment of inestial OF rod passing through the point of Suspension . Ir Ic+md? IR 7 Ic + neet my Gly? of Ic → Mac -- (Foo ood) M12 Ir t mce ---

now similarly passing through moment of inestia of sphere point be suspension be Is Ic + moment of inerta Of Sphere Passing througth center of spher da(2+0) azis of rotation for Is and I are As Parallel : by proauel axis theorem Is & Ic + md ² Is = 2m² + Mall+o)? Nii (Ic For Solid Sprere Ictamo) moment of inertia of the system I system a & IR + IS me2 + 3 m+ m (to)? - Vi (from ea

Let to be the final angular velocity of given system. le cohen o 70° • final rotation kinetic eneogy of system (R.KE), + 1/2 I systems wo? (R-ke), ☆ ☆ [moet + molt mo (Itnº je (R.KE) E + $ (mol + 3 mot mp (Atojº Joo? now on system only my is the force who is doning woook and my is conservative Force : Totat energy is conserved for given system

Final total Initial total energy (0-600) eneogy (070) final rotati Intial potential final potential energy to Initial Rotatio nal kinetic = energy t Kinetic energi energy nooo ooo U; + ' RIE; & Up + RKFF now from ego , , - og gif + mga unit) ] + 0 = male, t msg (ko) ]+ ek en .. Rrer + mogle mapt 4 mpg Clora)- og lang 2 RHEP a magulet megler) now from egs ® to [Mort? fm at ms Chtoj2Jc02= molt mo lang )

- co² mode + msg (ltr) 2 ☆ ( Mol?+ ? Moz? + Mg (1782) 4 2 W² a (mode + mr g letos] (mogle+ { Mg2+ my (1431] w? = Mode + msg (eta) molt 2 mg 82 tms (looja 20 Dow now that given mass of rod my ooky mass of sphere Ms 7 100 kg length of rod l = am radius of sphere =) 0.2 m as a 10 mija

co? 20 X10X 2 + 100X10X(240-2) 20%(2)2 + 2x 100 x 10.212 + 100x (2+0212 602 200 + 1000X22 20x%+ 0.4X100X0.04 + 100X@32 200 + 2200 20X1:33 + 1.6+489 w? 2400 26:67 +1.67484 co2 = 2400 S12.267 co 2 4.6850 16 2.164 road /sec Angularo velocity pendulum Tis 2.164 rad/see when 070°