power and force on a pendulum

Question

a pendulum of mass 1 kg and length 1 m is released from rest at angle 60 degree. The power delivered by all the forces acting on a bob at angle 30 degree is

Answer 1

Hi Aman!

 

Delivering power implies doing work and changing energy, since that is how power is defined.  For a pendulum, the only force doing work (and thus changing the pendulum's energy) is gravity.  The tension force that holds the pendulum on its circular path is a centripetal force, which is always perpendicular -- at each moment -- to the direction of travel, and thus does no work.

 

To get the power precisely at the moment when the pendulum, released from rest at 60 degrees (I will presume this is 60 degrees with respect to the vertical, since I am not told otherwise), is at 30 degrees from the vertical, we can use the formula:

 

P = Fv

 

where

 

P = power

F = force doing work

v = (linear) speed of object

 

If the bob is released from rest, we can calculate the speed v at the 30-degree point using a conservation of (mechanical) energy argument.  There would also be a way to calculate v using simple harmonic equations... but that 60-degree release angle is a little large for the pendulum's motion to be truly simple harmonic.  That is slightly over 1 radian, and we need the angles to be no larger, I would say, than a few tenths of a radian to have the pendulum's motion be more nearly simple harmonic.

 

So let's calculate v using (mechanical) energy conservation.  It turns out that, when a pendulum bob is held so that the pendulum makes an angle θ with the vertical, the vertical level (h) of the bob above its level at the lowest point in the swing is

 

h = L(1 - cosθ)

 

where L is the length of the pendulum.  Thus, for a 1-m pendulum, when θ = 60 degrees, we have

 

h₆₀ = (1 m)(1 - cos(60°)) = 1(1 - 0.5) = 0.5 m above lowest point in the swing

 

When θ = 30°, we have

 

h₃₀ = (1 m)(1 - cos(30°)) = 0.134 m

 

So, when the bob is released from rest at 60 degrees, it has (mechanical) energy:

 

ME₆₀ = KE₆₀ + GPE₆₀ 

 

KE₆₀ = kinetic energy at the start = 0 (since it is released from rest)

GPE₆₀ = gravtiational potential energy at 60 degrees (using the lowest level in the swing as h = 0)

           = mgh₆₀ = mg(0.5 m)

 

ME₃₀ = KE₃₀ + GPE₃₀ = (1/2)mv² + mgh₃₀ = (1/2)mv²+mg(0.134 m)

 

Since gravity is the only force doing work, (mechanical) energy is conserved.  Thus:

 

ME₆₀ = ME₃₀

mg(0.5 m) = (1/2)mv² +mg(0.134 m)

 

Notice that the mass cancels out of the equation when these two mechanical energy values are set equal.  We get:

 

0.5g = v²/2 + 0.134g

 

From there, you can solve for v, the speed of the bob when the pendulum is at 30 degrees.

 

Now, we need to determine the force of gravity on the bob at that point.  Actually, we only need the component of gravity in the bob's direction of motion (tangent to the circular path it is on), because this is the only component of gravity doing work on the bob (and thus delivering power to it).

 

In a very similar derivation to the one that calculates the component of gravity down an inclined plane of angle θ, it turns out the component of gravity in the bob's direction of motion is:

 

F_g-tangential = mgsinθ

 

This is tricky to describe without a diagram, but see if you can demonstrate it with a drawing.

 

Then:

 

P = Fv = F_g-tangential (v)

 

using the values calculated above.  This will be the power delivered to the bob at that instant.

 

I hope this helps some!  If you want to look more closely at the details, or check an answer, just let me know.