(a) In a Cartesian plane, if the position vector of a moving particle is given by F = ri+ yj, then the r and y components of velocity is given by and respectively. On the other hand, the polar components, i.e., the r and 0 components of velocity is given by and r. Using these, prove that the r component of acceleration is dr fr dt2 OP dt

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Chapter2: Second-order Linear Odes
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Please help me solving my calculus assignment. Please explain in details. Will be grately appreciated. 

3. Rotating Tube
(a) In a Cartesian plane, if the position vector of a moving particle is given
by i = ri+ yj, then the r and y components of velocity is given by
and # respectively. On the other hand, the polar components, i.e., the r
and 0 components of velocity is given by and r. Using these, prove
that the r component of acceleration is
d'r
fr
dt?
dt
(b) A smooth straight thin tube revolves with uniform angular velocity w in
a vertical plane about one end which is fixed. A particle inside the tube is
sliding along the tube with a constant velocity v. At time t = 0, the tube
was horizontal and the particle was at a distance a from the fixed end.
Show that the motion of the particle can be described by the differential
equation
dr
rw? = -g sinwt
dt?
[Hint: Use formula (10)|
(c) Find the complementary function (CF) and the particular integral (PI)
of the differential equation (11).
(d) Hence or otherwise, show that the general solution can be written as
r(t) = L cosh wt + M sinh wt +
sin wt
(e) Identify the initial conditions from the description in part (b) and use
them to find L and M. Hence show that
r(t) = a cosh wt +(-
-) sinh wt +
2w2
sin wt
2w2
Transcribed Image Text:3. Rotating Tube (a) In a Cartesian plane, if the position vector of a moving particle is given by i = ri+ yj, then the r and y components of velocity is given by and # respectively. On the other hand, the polar components, i.e., the r and 0 components of velocity is given by and r. Using these, prove that the r component of acceleration is d'r fr dt? dt (b) A smooth straight thin tube revolves with uniform angular velocity w in a vertical plane about one end which is fixed. A particle inside the tube is sliding along the tube with a constant velocity v. At time t = 0, the tube was horizontal and the particle was at a distance a from the fixed end. Show that the motion of the particle can be described by the differential equation dr rw? = -g sinwt dt? [Hint: Use formula (10)| (c) Find the complementary function (CF) and the particular integral (PI) of the differential equation (11). (d) Hence or otherwise, show that the general solution can be written as r(t) = L cosh wt + M sinh wt + sin wt (e) Identify the initial conditions from the description in part (b) and use them to find L and M. Hence show that r(t) = a cosh wt +(- -) sinh wt + 2w2 sin wt 2w2
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