Fowles and Cassiday Chapter 8: Planar motion of Rigid Bodies

Ph425 lecture notes (R. Egerton)

Planar motion implies that the axis of rotation is time-invariant.

8.1 Center of Mass

The coordinates of r_cm can be defined the same as for a system of particles.

But for a system with a large number of particles, the summations can be replaced by integrals involving the density r of the body, which is time-independent for a rigid body.

Often it is possible to assume that r is constant within the body.

For thin shells, the mass per unit area may be more convenient; or for thin rods, the mass per unit length.

If a body can be considered as a composite of two simple shapes whose C of M is easily known, the C of M coordinates are given by Eq.(8.1.5). (Example: letter L etc)

If the body has a symmetry plane or a center of symmetry, the C of M must lie on that plane or at that center. (Examples: letters V, X, C ; A)

8.2 Moment of Inertia

The C of M is involved in an object's static properties and its linear or orbital motion; moment of inertia affects its spin.

However, the spin axis need not pass through the C of M ; the moment of inertia depends of the body and the axis chosen.

Moment of inertia is defined as: I = S m_i r_i² where r_i is the distance of particle i from the axis of rotation, measured in a direction perpendicular to that axis.

If rotation is about the z-axis: r_i²= x_i² + y_i² .

I is the rotational equivalent of mass (in linear motion).

Thus we have (p.306): L = I w , torque N = I dw /dt , T_rot = (1/2) I w ² .

8.3 Calculation of I

For a rigid body, I_z = ò _r r² dm = ò ò ò _x,y,z (x² + y²) r dxdydz , (r = mass/vol)

Thin ring rotated about a central axis perpendicular to its plane: all the mass is at a distance a from the axis, so I = m a²

Thin rod rotated about an axis perpendicular to its length: I_z = ò x² r dx

where r = mass/length (incorporating integration over y and z)

If the axis is at one end of the rod, the limits of integration are 0 and a,

giving I = (1/3) r [(a)³ - (0)³] = (1/3)r a³ = (1/3) m a²

If the axis is halfway along the rod (at the C of M), the integration limits are -a/2 and a/2,

giving I = (1/3) r [(a/2)³ - (-a/2)³] = (1/12) r a³ = (1/12) m a²

If the axis is a distance b from the center, the limits are a/2+b and -a/2+b,

giving I = (1/3) r [(a/2-b)³ - (-a/2-b)³]

= (1/24) r a³ - (3/12) r a²b + (3/6) r ab²- (1/3) r b³

+ (1/24) r a³ + (3/12) r a²b + (3/6) r ab²+ (1/3) r b³

= (1/12) r a³ + r ab²

which has a minimum value at b = 0 (axis through C of M).

Disk is a two-dimensional example; dm = r 2p r dr where r is mass/area and integration is over radius r, from 0 to the disk radius a (p.309)

Cylinder is a 3-D example, but is equivalent to a set of disks aligned along the z-axis; integration over z simply converts r dz into (mass/area).

Sphere can also be considered as a set of disks, but disk radius is y(z) = (a²-z²)^1/2 and a single integration over z is required (p.309).

Spherical shell: I can be obtained by differentiating the result (expressed in terms of bulk density r ) for a sphere (p.309)

Represent x and y coordinates (relative to z-axis of rotation) as a sum of coordinates relative to the C of M and coordinates (x_cm,y_cm) of the C of M relative to the z-axis.

Then can show (p.312) that I_z = I_cm + m(x_cm² + y_cm²) = I_cm + m l²

where l is the perpendicular distance of the rotation axis from the C of M.

Example (p.313): coin rotating about an edge point: I_z = (m/2)a² + m a² = (3/2)ma²

I_z = S m(r_i²) = S m(x_i² + y_i²) = I_x + I_y

Only applies to a lamina in the (x,y) plane, otherwise r_i²= x_i² + y_i² + z_i²

Example: Coin spinning about a diametric axis: 2I_x = I_x + I_y = I_z = (m/2) a² , I_x = (m/4) a²

Coin rotated about a tangential x'-axis: I_x' = I_x + ml² = (m/4)a² + ma² = (5/4) ma² .

Convenient to set I = mk² , where k is the radius of gyration;

k depends on the shape of the object and the axis chosen, but is independent of the mass.

Values of k for simple objects are given in Table 8.3.1, p.315.

An object pivoted about an axis (distance l from the C of M) and swinging under gravity.

Can treat as orbital motion of the C of M, but the gravitational force is not central.

Torque N = - mgl sinq = I (d^2q /dt²) where l is the distance from pivot to C of M.

Same as for a simple pendulum (p.80) but with g/l replaced by mgl/I

If q is small, replace sinq by q , giving T = 2p [I/gml]^1/2

= 2p [(1/3)ma² / gm(a/2)]^1/2 = 2p [(2/3)a / g)]^1/2 for a thin uniform rod of length a = 2l , pivoted about one end - equivalent to a simple pendulum of length (2/3)a .

T = 2p [(I_cm+ml²)/gml]^1/2 = 2p [(k_cm² + l²)/gl]^1/2 = (2p /Ö g)[k_cm² /l + l ]^1/2

(Note: for a simple pendulum: k_cm = 0, giving T = 2p [l/g]^1/2 )

If l is varied, T becomes large for large or small l .

[k_cm² /l + l ] = A = (gT/2p )², l² - Al + k_cm² = 0

For a given T there are two solutions l and l', related by

k_cm² /l + l = k_cm² /l'+ l' giving k_cm² l' + l²l' = k_cm² l+l l'² ,

k_cm² (l - l') = ll'(l-l') , i.e. k_cm² = ll'

The alternative pivot point O' is the center of oscillation for point O, and vice versa.

The two solutions become equivalent if: l²= l'² = ll' = k_cm² , i.e. for l = k_cm.

This condition gives the minimum period of oscillation T_min for a given rigid body.

Uniform rod of length a: k_cm² = a²/12 .

If l = a/2, corresponding to pivot at one end, l' = a²/12l = a/6

For minimum T : l = k_cm = a/Ö 12 ,

T_min = 2p [(k_cm² + l²)/gl]^1/2 = 2p [(a²/6)(Ö 12/ga)]^1/2 = 2p [(0.577) a/g ]^1/2

In this case, the axis or rotation is not fixed but moves at constant velocity.

We can separate the motion into an orbital component (linear motion of the center of mass) and a spin component (rotation about the C of M). The force and torque are:

F = m a_cm , N_C = dL_C/dt = I_cm (dw /dt)

where a_cm is the linear acceleration of the center of mass

Example: a disk, cylinder or sphere rolling down a flat incline.

This general case is analyzed on pages 322 - 326.

It is convenient to take the x-axis as parallel to the inclined plane (Fig. 8.6.1).

Eqs.(8.6.4) represent the x- and y-components of the linear motion of the C of M.

Rotational motion is caused only by the frictional force F_P ,

so that dw /dt = F_Pa / I_cm where a is the radius of the object.

If contact friction prevents slipping, a_cm = d(a dw /dt) = a (dw /dt)

a_cm = a (F_Pa / I_cm) = a² F_P / I_cm

Combining this with Eq.(8.6.4a) for the linear x-acceleration gives:

a_cm = g sinq / (1 + k_cm²/a²) = constant (independent of time)

= (2/3) g sin q for a uniform solid cylinder (k_cm² = a²/2)

= (5/7) g sinq for a uniform solid sphere (k_cm² = 2a²/5)

Instead of considering forces, the above results can be obtained from conservation of energy. If there is no slipping, only mechanical KE and PE are involved:

Total energy E = KE + PE = (1/2)mv_cm² + (1/2)I(dw /dt)² + mgh_cm

Using w =v_cm/a , h_cm = -x_cm sinq and taking a time derivative gives

a_cm = g sinq / (1 + k_cm²/a²) as before.

If slipping occurs, F_P = m F_N = m mg sinq , where m = coefficient of friction

Using Eq.(8.6.4b) this time,

a_cm = g sin q - m g cos q = constant, v_cm = (sin q - m cos q ) g t if we take v_cm = 0 at t=0

Integration of the angular equation shows that v_cm = g (a w ) ,

where g = (k_cm/a)²(tanq /m - 1) .

g = 1 corresponds to no slipping, so the minimum coefficient of (static) friction needed to prevent slipping is: m = tanq / [1 + (k_cm/a)²]

A percussive force provides an impulse P = ò F dt

If applied to the center of mass of a free (unconstrained) body, this impulse induces only a change in linear velocity: P = m D v_cm .

If applied a distance l from the C of M, the force F also constitutes a torque N which induces rotation: ò N dt = Pl = I D w

Example: collision of a baseball with a bat (p.329)

Treat the ball as a particle, mass m, and imagine the bat (mass M) to be unconstrained (e.g. resting on a smooth horizontal surface).

P is the impulse delivered from the ball to the bat.

For the ball: -P = mv₁ – mv₀

For the bat: P = Mv_cm – 0 and

Pl’ = I_cm w – 0 if P acts at O’, a distance l’ from the C of M (Fig. 8.7.1)

For an arbitary point O, a distance l from the C of M and on the opposite side from O’

v_O = v_cm - w l = P/M – (Pl’/ I_cm)l

= 0 if 1/M – (ll’/ I_cm) i.e. ll’ = I_cm/M = k_cm²

This defines the condition for O being the center of percussion of an object (radius of gyration k_cm) for the impulse P.

If the baseball is approximated as a uniform rod of length 2l, hinged at one end,

(k_cm)² = (2l)²/12 = l²/3 , giving l’ = l/3 ,

l+l’=(4/3)l =(2/3)(2l) i.e. the bat should be struck at a point one third from the end.

The conclusion is the same if the bat is swung and the ball is stationary, or if both have motion prior to the collision.

An equivalent example is a door, which (viewed from above) approximates to a uniform rod. If the door is swung, a doorstop placed at the center of percussion will result in all the angular momentum being absorbed by the stop and no force on the hinges. (However, the stop should be at one half the door height, otherwise the impulse will produce a torque about a horizontal axis).