The pendulum shown has mass 2.5kg and is released from rest when it is in the ho

Question

The pendulum shown has mass 2.5kg and is released from rest when it is in the horizontal position (top). The rod is negligible and the mass can be treated as small (negligible radius).

a) Calculate the linear speed of the mass at the other two positions shown in the diagram. Do this by considering the rotational energy of the system instead of linear. (Of course, it gives the same answer both ways.)

b) If the rod has mass M (known) and length L (also known), recalculate the quantity in part a.

Explanation / Answer

a)

At the first point shown, the bob has fallen through height
h = Lcos
and so has lost PE = mgh = mgLcos

At the second point shown, h = L
and the PE loss is mgL.

The rotational kinetic energy is given as
KE = ½I²
and for a point mass m a distance L from the pivot
I = mL², so
KE = ½mL²²

So for the first case
mgLcos = ½mL²² divide by mL²/2
² = 2gcos / L
and when I plug in g = 9.8m/s² and = 30º and L = 2.00m I get
² = 8.5 rad/s²
= 2.9 rad/s
and since v = *L,
v = 2.9rad/s * 2.00m = 5.8 m/s speed at 30º

For the second case
mgL = ½mL²² leads to
² = 2g / L leads to
² = 9.8 rad/s²
= 3.1 rad/s
v = 6.3 m/s speed at bottom

b) Now in the first case the bob has fallen through h = Lcos
but the CM of the rod has fallen through h/2 = Lcos / 2
so the total loss of PE is
g(m + M/2)Lcos.

For the second case, we lose the trig function, and the PE loss is g(m + M/2)L.

The new moment of inertia of the assembly is
I = mL² + ML²/3 where M is the mass of the rod, not given

Therefore for the first case
g(m + M/2)Lcos = ½(m + M/3)L²*²

Unlike in part (a), this time I'm going to substitute for straight away:
² = (v / L)², so
L²² = L² * (v / L)² = v², so we have
g(m + M/2)Lcos = ½(m + M/3)v²
v² = 2gL(m + M/2)cos / (m + M/3)
If you know M, you can get a numerical result for v² and then v.

For the second case (pendulum vertical), we get the same result without the trig function:
v² = 2gL(m + M/2) / (m + M/3)

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