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Classical Mechanics March 2009 Exercise sheet 4

Exercise 1

Calculate the moment of inertia of a homogeneous cone spining about its axis ofsymmetry;

Calculate the moment of inertia of a homogeneous disc rotating about a diameter;

Calculate the moment of inertia of a homogeneous disc rotating about an axis parallelto a diameter, and passing through the axis of symmetry which is perpendicular tothe disc;

Calculate the moment of inertia of a homogeneous cone spining about the axis per-

pendicular to its axis of symmetry, and which contains the tip of the cone.

Exercise 2A wheel of moment of inertia I with respect to its axis of symmetry rotates around under the influence of a constant torque and a friction torque proportional to theangular velocity , where is a constant. Write the equation of motion of the wheel andgive the solution corresponding to (0) = 0.

Exercise 3A vertical pulley wheel of radius R can rotate around its axis of symmetry with momentof inertia I. A rope of negligible mass is wound around the pulley and has a mass m tied

at the other end. If the rope does not slip on the pulley wheel, use energy conservation toderive an expression for the angular acceleration of the pulley.

Exercise 4 (exam 2006 - section B)A pendulum is made out of a homogeneous rod of length l and mass m, and a homogeneousdisc of radius R and mass M, whose centre is fixed to one end of the rod. The other end ofthe rod is fixed to a horizontal axis perpendicular to the rod, around which the pendulumcan oscillate.a) Show that the moment of inertia of the rod about the axis is 1

3ml2.

b) Derive an expression for the moment of inertia of the disc with respect to a perpendicular

axis passing through its centre. Using the appropriate theorem, show that the moment ofinertia of the disc about is MR2

2+ l2

.

c) If the angle between the pendulum and the vertical is , show that the potential energy

of the pendulum is U = glm2

+ M

(1 cos ), if the zero of the potential is at = 0.

d) Using energy conservation, show that the equation describing the oscillations is

+3gl(m + 2M)

2ml2 + 3M R2 + 6M l2sin = 0.

e) Under what condition does this last equation describe harmonic oscillations? Obtainan expression for the angular frequency of these oscillations.

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Solutions

The mass density is denoted by , the hight of the cone is H, its mass M and its radiusat the base R.

The cone can be seen as a collection of discs of radius r(z), z varying from 0 to H,each with moment of inertia (1/2)dmr2(z) (see lecture notes), where dm = r2(z)dzand r(z) = Rz/H. The moment of inertia of the cone is then

I1 =1

2

R4

H4

H0

dz r4(z) =1

10R4H.

The mass of the cone is

M =H0

dm = H0

dz r2(z) =1

3R2H,

such that

I1 =3

10MR2.

The disc of radius r can be seen as a collection of stripes of lenght l(z) and thicknessdz, with l2(z) = 2(r2 z2), and z varying from 0 to r. If is the mass per unitsurface area, the moment of inertia of a stripe is

dI2 = dzl/2l/2

dxx2 =1

12l3(z)dz =

1

12dml2(z).

The moment of inertia of the disc in this motion is then

I2 = 2r0

1

12l3(z)dz =

4

3

r0

(r2 z2)3/2dz,

where the change of variable z = r sin , for [0, /2], leads to

I2 =4

3r4

/2

0

cos4 d.

Using the complex exponential representation for the cosine, we find

cos4 =3

8+

1

2cos(2) +

1

8cos(4),

such that

I2 =1

4r4 =

1

4mr2,

where m = r2 is the mass of the disc.

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Using the parallel axis theorem, the moment of inertial with respect to an axis abovethe disc and parallel to a diameter, at the distance z, is

I3 = I2 + mz2 =

1

4r2 + z2

m

From the previous result, the moment of inertia of the cone about the axis perpen-dicular to its axis of symmetry, and which contains the tip of the cone, is

I4 =H0

1

4r2(z) + z2

dm(z),

where dm(z) = r2(z)dz and r(z) = Rz/H. The moment of inertial is therefore

I4 =35

M

H2 +

R2

4

2. The equation of motion is I = , and the most general solution is

(t) = 0 exp

t

I

+

,

where 0 is a constant of integration. With the initial condition (0) = 0, one obtainsthen

(t) =

1 exp

t

I

.

3. The total energy of the system is

E =1

2I2 +

1

2m(z)2 + mgz,

where z is the height of the point mass. Because the rope doesnt slip, we have z = R,and therefore energy conservation 0 = E = I + mR2 + mgR leads to

=mgR

I+ mR2,

after a division by = 0. (Note the sign convention: > 0 if the pulley rotates anticlock-wise)

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4. a) By definition, the moment of inertia is

dm r2, where r is the distance of theinfinitesimal mass dm to the axis of rotation, and the integral runs over the volume of the

rigid body. This gives here

I1 =l0

x2dx = l3

3=

1

3ml2,

where is the mass density and m = l the mass of the rod.b) The moment of inertia of the disc with respect to the axis of symmetry perpendicularto the disc and passing through its centre is

Ro

rdr

2

0

dr2 =

2R4 =

1

2MR2,

where M = R2 is the mass of the disc. The Parallel Axis Theorem gives then themoment of inertia with respect to as:

I2 = M l2 +

1

2MR2.

c) The heights are measured from the position = 0. The height of the centre of the discis l(1 cos ) and its potential energy is then M gl(1 cos ). The height of the centre ofthe rod is l/2(1 cos ) and its potential energy is then mgl/2(1 cos ).d) The kinetic energy of the system is 1/2(I1 + I2)()2, and energy conservation reads

ddt

12

(I1 + I2)()2 +

M gl + mg l

2

(1 cos )

= 0

which leads to the expected equation of motion, after a division by = 0.e) For small amplitudes