Efficiency of energy transfer on a rolling object

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Holly O’Nione 11HO

 Aim: To investigate the efficiency of energy transfer on a rolling object.

To do this I will investigate the input variable which is the factor we’re changing. I shall change the height of the slope, which therefore means the input variable is the gravitational potential energy (G.P.E) as the height moved affects it.

Prediction

I predict that the transfer of energy will be more efficient the steeper the slope gets. This is because as the slope gets steeper, the marble will rotate less and less (so less rotational energy) and make less and less contact , and when the slope has a gradient of 100%, the marble will not rotate at all, but just fall straight down.  

The marble’s energy changes from G.P.E, as it falls vertically, to kinetic energy. However, as it comes down the slope it rotates. Some energy is lost as rotational energy.

G.P.E = K.E + Rotational Energy

Therefore I will need to measure the velocity at the bottom as there is no Kinetic Energy (K.E) or rotational energy is at the top.

I will need to find out the efficiency of the energy transfer from G.P.E to K.E. The formula is:

Efficiency = Useful energy transferred (KE) x 100

                   Total energy supplied (GPE)

To work this out I need to know the G.P.E the marble has at the top (therefore the maximum G.P.E) and the amount of K.E the marble has at the end of the slope (therefore the max. K.E).

We can find G.P.E by using the following formula:

GPE = mgh

m = the mass of the object in Kilograms

g = the acceleration due to gravity in metres per second squared

h = the height in metres.  

We can find out K.E. by using the following formula:

KE = ½ mv2

m = the mass of the object in Kilograms

v = the speed of the object in metres per second.  

To work out the average speed:

Avg. speed =_x_ = u + v

                 t         2

x = the length of the slope in metres

t = the time taken for the marble to roll down the slope in seconds

u = the initial speed of the marble

v = the final speed of the marble.

But as the marble will be stationary u = 0, so this means:

_v_ = _x_

  2         t 

If we bring 2 over to the other side this gives us a formula to work out the speed:

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v = _2x_

          t

So to work out the efficiency the formula can be rewritten as:

Efficiency = ½mv2 x100

                      mgh

The mass cancels out and so the mass of the marble does not have to be recorded.

Efficiency = ½v2 x100

                       gh

If I sub the formula for v I get the following formula:

Efficiency = 0.5(2x/t)2     x 100

                       gh

I shall use this formula as it requires only one calculation and still ends ...

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