Taylor 1.7 (today) and 2.1-2.2 (Friday)
We showed that application of Newton's third law leads to:
If the net external force Fnet on an N-particle system is zero, the system's total momentum P is constant.
A block of mass m is observed accelerating from rest down an incline that has coefficient of friction μ and is at angle θ from the horizontal. How far will it travel in time t?
First, draw a diagram of the problem.
Second, choose a coordinate system. For problems of this type, the simplest system has the x axis along the incline, the y axis perpendicular to the incline, and the z axis across the incline.
Third, resolve the forces along the 3 axes. Along z the force is zero, so the acceleration in this direction is zero and the velocity is constant. Since the block is initially at rest, there is no motion along z.
The block remains on the incline, so the initial zero velocity in the y direction must remain zero, meaning the net force in the y direction must be zero. The forces in the y direction consist of the component of the weight perpendicular to the incline, mg cosθ downward, and the normal force of the incline against the block, N upward such that N=mg cosθ.
The block can move along the x direction, so we don't require or expect the net force to be zero. We have the component of the weight along the incline, mg sinθ tending to move the block down the incline, and the frictional force acting against it. Recall that the common approximation for the force if friction is f=μN. Here we will assume that there is only a kinetic friction force and ignore the difference with static friction. From the constraint of the forces in the y direction we know the normal force in terms of the weigh and angle of incline, and can use that to write down the equation of motion in the x direction:
We can solve this by canceling the common mass factor and integrating twice with respect to time to get
Although they seem to add unnecessary complexity, polar (or cylindrical or spherical) coordinates can make some problems significantly easier to solve. The added complexity comes mainly from the fact that the unit vectors are not fixed, so derivatives of quantities must include terms for the change of the unit vectors.
Polar coordinates and Cartesian coordinates are related through:
How do we define the unit vectors rhat and φhat? This is easy to think about if you realize that the unit vector xhat is chosen to point in the direction of incresing x with y held constant, and likewise for yhat. Following this idea, we define rhat as the direction of increasing r with φ held constant, as shown in Fig. 1.11.
An alternative definition is rhat = r / |r|. This definition works in general, for any vector a, ahat = a / |a|.
φhat is perpendicular to rhat. Any vector can be decomposed into orthoganal pieces in the rhat and φhat directions, for instance, any force vector can be written as:
The position vector in polar coordinates is r = r rhat. Therefore, Newton's second law reads simply F = mr[ddot]. We need to work out r[ddot] in polar coordinates.
First, notice that since rhat and φhat are unit vectors, their lengths are fixed; any change to a unit vector must be in a direction perpendicular to the vector. Make a sketch of some arbitrary motion in two dimensions and pick two nearby points with slightly different rhat vectors, r1hat and r2hat. Call the angle between the two vectors Δφ, and notice that Δrhat = r2hat - r1hat ≈ Δ&phi φhat. Let the time it takes for the object to move from point 1 to point 2 be Δt. Then
Similarly we can show that
Now
The second derivative of r is
Now we're ready to solve Newton's law in polar coordinates.