# Advanced Mechanics (Set 1)

1. A particle experiences two forces. One constant force is given as F = F_0 x(^) (vector F = F(initial) * x(unit vector)) and the other force is a drag force that is proportional to the velocity with proportionality constant ß.Write out Newton’s 2nd Law for this particle and solve the differential equation for the particle’s position along the x-axis as a function of time. Show that if the particle starts from rest, the velocity will reach a limiting value of F_0/ß.

2. A force is given by: (vector)F(x, y, z) = (yz)x(^) + (xz)y(^) + (yx)z(^).

a.) Find the work done by this force when a particle is taken from (0, 0, 2) to (2, 4, -3). Can you choose any path, or does the answer depend on the path?

b.) Can a potential energy function be specified for this force? If so, find an appropriate function.

3. A solid cylinder of radius b, mass m, rolls down an incline of angle ß. Use a coordinate system attached to the lower end of the incline and find the acceleration of the cylinder along the plane.

1. The movement is along x axis. Both forces are along x axis

$F1 =F0$

$F2 = \beta*(d x/d t)$

$F =m*(d^2x/dx^2)$

$F0- \beta*(d x/d t) -m*(d^2x/dt^2) = 0$ (2)

or

$m*(d^2y/dx^2) + \beta(d y/d x) = F0$ (2)

the homogeneous equation is

$+m*(d^2y/dx^2) +\beta(d x/d y) + 0*y = 0$ (1) the characteristic equation attached is

$m*z^2 +\beta*z + 0 =0$

$z(m*z+\beta) = 0$

$z1 =0$

$z2 = -\beta/m$

the solutions of the homogeneous differential equation are

$y1(x) = C1*exp(0*x) = C1$

$y2(x) = C2*exp(-\beta*x/m)$

So the general solution of the homogeneous equation (1) is

$Y(x) = C1+ C2*exp(-\beta*x/m)$

For the equation (2) since the free term F0 is constant the the general solution is

$Y(x) = C1 +C2*exp(-\beta*x/m) + a*x$

Now one has

$\beta*(d y/d t) +m*(d2y/dt2) = F0$ (2)

$d y/d x = – \beta/m*C2*exp(-\beta*x/m) + a$

$d^2y/dx^2 = (\beta/m)^2*C2*exp(-\beta*x/m)$

$(\beta^2/m)*C2*exp(-\beta*x/m) +a + \beta^2/m*C2*exp(-\beta*x/m) = F0$

$a = F0$

Hence

$d y/d x = -(\beta/m)*C2*exp(-\beta*x/m) – F0$

which at the limit $x \rightarrow \infty$ is

$d y/d x = F0$

2.

$F = yz * i +xz* j + xy * k$

where i, j, k are the unity vectors along x,y,z axes

$dr = i*dx +j*dy+k*dz$

hence the infinitesimal work is (i*j =0, i*k =0, j*k=0)

$dW = F*dr = yz*dx + xz*dy + xy*dz$

$W = \int _0^2 (yz*dx) + \int _0^4 (xz*dy)+ \int_{-3}^2 (xy*dz) =$

$= yzx (x from 0 to 2) + xzy(y from 0 to 4) +xyz (z from 2 to -3) = 2yz + 4xz – 5zy$

You can choose any path to calculate the total work as long as the initial and final points are the same hence it is conservative.

The force F is conservative if it exists a scalar field E such as $F = \nabla (E)$.

In our case $E = x*y*z$ is the potential energy of the field.

3. Let a system xy of coordinates attached to the lower end of the inclined. Let y0 be initial height of the cylinder on the inclined plane.

the moment of inertia of a solid cylinder having radius b and mass m is $I =m b^2/2$

Now at the base of the inclined plane one has the total energy (rotation + kinetic)

$E = I*\omega^2/2 + m*v^2/2$

which need to be equal to the initial potential energy $E =m*g*Y0$

now one has for the points of cylinder in contact with the inclined $V1= 0$, for the points of the cylinder on the axis V and for the points opposite to the contact point the speed $V2= 2*V$

Hence $\omega V = \omega*b$

$I*\omega^2/2 +m*\omega^2*b^2/2 = m*g*Y0$

$\omega^2/2 *(I+mb^2) =m*g*Y0$

$\omega^2/2 *3mb^2/2 =m*g*Y0$

$\omega^2 = (4*g*Y0/3b^2)$

$V^2/b^2 = 4*g*Y0/3b^2$

$V = \sqrt{4gY0/3}$

now the length of the inclined plane is $L =Y0/sin(\beta)$

$V^2 = 2*a*L$ where a is the acceleration along the plane

$4 g Y0/3 =2*a*Y0/sin(\beta)$

$a= 2*g*sin(\beta)/3$

4.speed of particle is $V = V0 * i$

Force is $F = q E*k + q(V x B) = q E*k + q V B (i x j) = q E*k +q v B*k$

where i is along x, j is along y and k is along z

hence $F= Fz = q E + q v B$

the equation of motion are

$m*d^2x/dt^2 =0$

$m*d^2y/dt^2 =0$

$m*d^2z/d^t2 = Fz$

$dx/dt = C1$, $C1 =V0$

$x = V0*t + C2$, at $t=0$ $x=0$, hence $C2 =0$

$d y/d t = C3$, $C3 =0$

$y = C4$, at $t=0$, $y =0$, hence $C4 =0$

$dz/d t = Fz*t + C5$, at $t =0$, $dz/dt =0$ hence $C5 = 0$

$z = Fz*t^2/2 + C6$, at $t=0$, $z =0$ hence $C6 =0$

$z =(q E+q v B)*t^2/2$