Taylor 1.6 - 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 and equal in magnitude to the downward component of the weight 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 kinetic friction and ignore that it can differ from static friction. From the constraint of the forces in the y direction we know the normal force in terms of the weight 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
A and B are constants of integration. We find their values from the initial conditions: the block starts from rest at t = 0, and we wish to find the distance traveled from the initial point, so let x = 0 at t = 0 and the distance traveled at time t is given by x(t). The first condition yields A = 0, and the second yields B = 0, so the final answer is
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:
This could represent the force on an object being twirled on a string, with Fr equal to the tension in the string, and Fφ equal to the force of air resistance on the object.
The position vector in polar coordinates is r = r rhat. Therefore, Newton's second law reads simply F = m d²tr. 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
Notice that this is the familiar expression for the radial and tangential components of acceleration.
Now we're ready to solve Newton's law in polar coordinates.
Consider the simple pendulum problem in polar coordinates, that is a mass at the end of a string, and solve for the motion for small oscillations.
Put the center of the coordinate system at the fixed end of the string and let φ=0 at the equilibrium position where the pendulum hangs vertically. There are two forces acting on the pendulum, the tension in the string, and the weight of the mass. The components of the force are Fr = mg cosφ and Fφ = -mg sinφ. The tension in the string will be equal and opposite to the radial component of the force, T = -Fr yielding no motion in the radial direction.
In the φ direction we have
Since there is no motion in the r direction, the second term on the right is zero, leaving
For small amplitudes, sinφ ≈ φ and the equation of motion becomes
This is a differential equation, a linear, homogeneous second order differential equation to be precise. This differential equation can't be solved by simple integration, but we can "guess" the form of the solution. It says that φ is a function of time whose second derivative with respect to time returns the same dependence on time. For instance, if φ(t) is a polynomial of order n, then the second derivative is a polynomial of order n-2, so this can't be the functional form for φ. What functions do you know of with second derivatives that are the same as the original function? The sin and cos functions have this property. So we try a solution of the form φ(t) = A sin(ωt) + B cos(ωt)
Use the initial conditions to find the specific solution.