Chapter 10, Problem 45 GO Gradebook of a simple pendulum is 0.65 m and the mass

Question

Chapter 10, Problem 45 GO Gradebook of a simple pendulum is 0.65 m and the mass of the particle (the "bob") at the end of the cable is 0.39 kg. The pendulum is pulled away from its equilibrium position by an angle of 8.2 and reieased from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth (c) What is the bob's speed as it passes through the lowest point of the swing? ron e s

Explanation / Answer

Height from the reference achieved by the bob after pulling = 0.65 x cos(90-8.2)0 = 0.093 m

Lets find the Time to taken reach the mean position :

S = ut + (1/2)a.t2 . Here u = 0, a = 9.81 m/s2, s = 0.093m, t=?

We can calc t= 0.138 sec

Hence total time taken to complete one cycle (in SHM) = 4x0.138 = 0.55 sec

a) Angular freq = 2(pi)/T = 2x(pi)/0.55 = 11.42 rad/sec

b) Lets consider h is the height from the lowest position, Potential energy of the bob at top position = mxgxh = 0.39x9.81x0.093 = 0.356 J

The total mechanical energy is the sum of PE and KE at any point.

At the top point, there is no KE as the ball is at rest, so the PE is the total mechanical energy. This PE gets converted to KE as the bob moves downward.

So, mechanical energy of the pendulum is 0.356 J

c) The potential energy at top position gets converted to KE at the lowest position and there is no PE there.

Hence, (1/2)mv2 = 0.356

i.e. v = 1.35 m/s ==> this is the speed at the lowest position

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