class 051010

Fill in the following information for the experiment you conducted Friday:

Sketch the weight vector and the normal-force vector for a ball rolling down a inclines of 15 degrees, 45 degrees and 75 degrees.

For each incline sketch the parallelogram formed by these two vectors and the resultant of the two vectors.

Key characteristic of your pictures:

• Presumably the ball doesn't change mass as the incline changes, so the weight vector should be the same for all inclines.
• The net force, the resultant across the diagonal of the parallelogram, will be in the direction down the incline, since we know that the acceleration is along the incline.
• If we draw a line from the end of the weight vector to the incline in a direction perpendicular to the incline, that segment will be one side of the parallelogram and will therefore be parallel to and of the same length as the normal force vector.

If frictional force was included:

• The frictional force for a coasting object is proportional to the normal force.  More normal force, more frictional force.

If the PE of a ball rolling down an incline decreases by .65 Joules while the ball does a total of .05 Joules of work against dissipative forces, then by how much does its KE increase?

`dW_net_on = `dKE.

`dW_net_on = `dW_cons_on + `dW_noncons_on

`dPE = `dW_cons_by = -`dW_cons_on

(work done by the system against conservative forces is equal equal and opposite to the work done on the system by conservative forces.

This is because the force exerted by the system against conservative forces is equal and opposite to the force exerted on the system by those conservative forces.

Standard example:  If I lift 200 Newtons thru a displacements of 2 meters upward then I do positive work against gravity:  2 meters * 200 Newtons = 400 Joules.  Gravity is acting downward so the work it does it 2 meters * (-200 N) = -400 J).

So

`dW_net_on =

`dW_net_cons_on + `dW_net_noncons_on =

-`dW_net_cons_by - `dW_net_noncons_by =

-`dPE - `dW_net_noncons_by

So

-`dPE - `dW_noncons_by = `dKE,

where `dW_noncons_by is the work done by the system against nonconservative forces.

A decrease in PE tends to increase KE, and negative work done against nonconservative forces tends to decrease KE.

If the ball has mass .1 kg, then

• what is its vertical drop on the incline?
• if it is released from rest, what velocity will it attain?

Experiment analysis:

The final KE of the ball in the experiment is not just .5 m vf^2.

The final KE is .5 m vf^2 + .5 I (omega_f)^2, where

• I = 2/5 m R^2, m being the mass and R the radius of the ball and
• omega = v / R (velocity of the ball divided by its radius)

Find the final velocities of the balls, based on your measurements of range.

Find change in KE, using the formula given above.  All you need to use that formula is mass, final velocity and radius.  Assume a radius of 2 cm and mass .1 kg.