The length of a simple pendulum is 0.81 mand the mass of the particle (the \"bob

Question

The length of a simple pendulum is 0.81 mand the mass of the particle (the "bob") at the end of the cable is0.23 kg. The pendulum is pulled away fromits equilibrium position by an angle of 8.10° and released from rest. Assume that frictioncan be neglected and that the resulting oscillatory motion issimple harmonic motion. (a) What is the angular frequency of themotion?
1(No Response) rad/s

(b) Using the position of the bob at its lowest point as thereference level, determine the total mechanical energy of thependulum as it swings back and forth.
2(No Response) J

(c) What is the bob's speed as it passes through the lowest pointof the swing?
3(No Response) m/s (a) What is the angular frequency of themotion?
1(No Response) rad/s

(b) Using the position of the bob at its lowest point as thereference level, determine the total mechanical energy of thependulum as it swings back and forth.
2(No Response) J

(c) What is the bob's speed as it passes through the lowest pointof the swing?
3(No Response) m/s

Explanation / Answer

The Time period of the pendulum        T = 2 [ L/g ]           = 2 [0.81 /9.8 ]           =1.8s The angular frequency is         = 2 / T             =2 / 1.8            = 3.48 rad/ s --------------------------------------------------------------------------------------------------------- The height through which the bob is lifted up is         h = L ( 1 - cos )            =0.81 ( 1 - cos 8.1 )            =8.08x10-3 m The total mechanical energy is         E = m g h            =0.23(9.8)8.08x10-3 m            =1.8214x10-2J ------------------------------------------------------------------------------------------------------------- When the bob passes through the lower position, the energy iscompleted in the form of kinetic energy       (1/2)mV2 =1.8214x10-2J              V = 0.3979m/s                  = 0.4m/s

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