What is the speed of the pendulum at the lowest point in the following case: A pendulum is swinging so that its bob just misses the floor at its lowest point and its highest point (at the two positions just before it changes direction) is 1 m off the ground.

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Alternatively we can use the conversion of gravitational potential energy (maximum at top of swing with minimum kinetic energy because it is momentarily motionless) to kinetic energy at the bottom of the swing (maximum because it is moving at its fastest here with minimum gravitational potential energy because it can't fall any lower).

From the law of conservation of energy:

E potential lost  =  E kinetic gained

hence      mgh  = 0.5mv^2    divide both sides by m

gh  =  0.5v^2      rearrange for v

v   =  (2gh)^0.5  =  (2  x  9.8  x  1)^0.5  =  4.43ms^-1

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The speed of the pendulum at its highest point is zero. As it starts to move downwards, the speed increases and reaches the maximum when it is at its lowest point. This gives the pendulum the maximum kinetic energy at the lowest point. Also, its potential energy is zero here. When the pendulum starts to climb up again, the kinetic energy is converted to potential energy.

In the question, the pendulum has a highest point at a height of 1m. The acceleration due to the gravitational force of the Earth is g= 9.8 m/s^2.

Now we use these formula between distance travelled, speed and acceleration:

a = (v - u)/t

=> t = (v- u)/a

d = (1/2)(v + u)*t

=> d = (1/2)(v + u)(v - u) / a

=> d = (v^2 - u^2) / 2a

=> v^2 = 2ad + u^2

Now we have d = 1 m , u = 0 m/s , a = 9.8 m/s^2. Substituting these values in the formula gives us:

v^2 = 2* 1* 9.8 + 0

=> v = sqrt ( 19.6)

=> v = 4.427 m/s

Therefore the required speed at the lowest point is 4.427 m/s.

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