568. Let a, b, be the lengths of the arms of the lever; A, B their weights, and a = the between them. Then since when in equilibrio, the vertical line passing through the point of suspension also passes through their centre of gravity, we easily get the distance (k) of this centre of gravity from the axis of sion, viz.

k =

a sin. B 26B sin. a

× √ {1 + (aA + bB cos: a)3}

suspen

where ẞ denotes the angle contained by a, and the line joining the extremities of the lever.

Again, the mass is

M = A + B.

Again, if x denote the distance of any point in the arm a from the point of suspension, then the momentum of inertia of that point is

Similarly it may be shewn that

B3 sin.3 (a + d).

is the momentum of inertia of the other arm.

Hence the distance of the centre of oscillation from the point of suspension is (Venturoli).

the length of the simple pendulum. But the time of an oscillation in the simple pendulum whose length is 1, (see Venturoli) is

569.

Let w be the given weight of the rod, a its length, and T the given time of oscillation; then the distance of the centre of oscillation from that of suspension is

r being the distance of the centre of gravity of the rod from the point of suspension. Hence

570.

Let w be the given weight of the cylinder, r the radius of its base, and b its length; also let w' be the weight of the chain, its whole length, and a that of the part thereof un

wound at the commencement of the motion; then since the inertia

and the moving force after the space x has been described

Let xa; then the time of unwinding the length - a is

a

571. If v,

denote the velocities of the oscillating and

revolving bodies at the distance from the lowest point of the circle, then we have

v = √2g.(h−x), v′ = √↓ 2y.(h' — x).

s mea

Again, let t, t' denote the times in which the arc 2πl sured from the highest point, s being that part of the semicircle which corresponds to the abscissa x; then

4g

2 21

To find the velocity V and periodic time T in the circle, not considering the accelerations due to gravity, we have

Hence moreover since the periodic time in a circle is generally

R being the radius and F the centripetal force, the centripetal force in this case, or the tension of the string is,

572.

Let A, B, be the weights of bodies, B being that of the one projected; also let a be the distance of B from the ring before projection, and its distance after it has been in motion during the time t, and make the angle between and a, and Then, if u denote the velocity

let ß be the velocity of projection.