CHAPTER IX.

MOTION OF RIGID BODIES. ROUGH SURFACES.

SECT. 1. Single Body.

(1) A CYLINDER descends down a perfectly rough inclined plane by the action of gravity, its axis being horizontal: to determine the motion of the cylinder and the friction at any time of its descent.

Let G (fig. 207) be the centre of gravity of the cylinder at any instant of its descent; 04 the course of the point of contact H of the circular section of the cylinder through G down the inclined plane; let OH = x, a = the angle of inclination of OA to the horizon, 0= the whole angle through which the cylinder has revolved about its centre of gravity in moving from 0 to H; a= the radius of the cylinder, k = its radius of gyration about its axis. Let F denote the friction of the plane on the cylinder, which, from the signification of perfect roughness, is supposed to be sufficient to prevent sliding, and m the mass of the cylinder.

Then for the motion of the cylinder we have, resolving forces parallel to OA,

But, since F is sufficiently great to secure perfect rolling, it

is evident that x=a0; and therefore, by (2),

(2) A globe descends from instantaneous rest down the surface of a perfectly rough hemispherical bowl, the centre of the globe always remaining in the same vertical plane: to determine the velocity of the globe at any position of its descent.

Let ABA' (fig. 208) be the vertical section of the bowl made by the plane in which the centre C of the globe is always situated, O being the centre of the bowl and OA a horizontal radius. Let M be the point at which the globe touches the bowl at any time of its motion, B being the initial position of M. Draw the radii OB, OCM; and let C'C be the circular arc described by the centre of gravity of the globe.

=

Let

< AOM=0, AOB = a, C'C=s, a = the radius of the globe, r = the radius of the bowl, & the angle which the globe has described about its centre of gravity in the motion from B to M, m = the mass of the globe; F= the friction of the bowl upon the globe at the point M, which is supposed to be sufficiently great to prevent all sliding.

Then for the motion of the centre of gravity of the globe, which will not be affected by our supposing all the impressed forces to be applied at C in their proper directions,

and, for the motion of the globe about its centre of gravity,

From the points B and M, draw two indefinite straight lines BkB' and MkT, tangents to the section of the hemispherical bowl; along BB' measure a length Bm equal to the circular arc BM; then, if we were to conceive the globe to roll from B along the length Bm, and then Bm to be applied along BM so as to coincide with it, mB' being, as soon as m coincides with M, a tangent both to the circle AMA' and to the globe; it is evident that the globe would have revolved about its centre through the same angle as by its actual motion of rolling down the arc BM. Now by rolling along Bm it would have Bm BM p

revolved about its centre through an angle

4

a

=

a

1 (0-a);

=

a

and, by the transference of m to M, it would have revolved through an angle equal to < B'kM = ▲ BOM=0-a, in an opposite direction. Hence we see that the whole actual angle through which the globe revolves about its centre in its actual

motion from B to M, is equal to (0 − a) = $.

-α
a

Hence, putting for 4 its value in (2), we have

Again, it is clear from the geometry that s=(r — a) (0 − a), and therefore, from (1),

Eliminating F between (3) and (4), we obtain

For the magnitude of the friction at any time, we have, from (3),

(3) A heterogeneous sphere rolls along a perfectly rough horizontal plane, its rotatory motion taking place always about an instantaneous axis normal to the vertical plane which passes through its geometrical centre and its centre of gravity: to determine its angular velocity for any position in its path.

Let C (fig. 209) be the geometrical centre and G the centre of gravity of the sphere at any time; S the point of contact of the vertical section of the sphere containing C and G with the horizontal plane; OSE the rectilinear locus of the points of

contact; CGA a radius of the sphere; GM, CS, perpendiculars upon the plane.

Let Fdenote the friction of the plane at any time upon the sphere, estimated in the direction EO, and R the vertical reaction of the plane; let m = the mass of the sphere; k = the radius of gyration about an axis through G at right angles to the vertical section containing C and G; OM = x, GM=y, CS CA=a, AGM ACS=6, CG = c.

=

Then for the motion of the sphere we have, resolving forces parallel to OE,

and, taking moments about the centre of gravity,

Now, since the friction is supposed to be sufficiently rough to prevent all sliding, we have from the geometry,

x+csin &=b+ap,

b being the value of x when = 0; and therefore