Page 70 - Mechanical Engineer's Data Handbook

APPLIED MECHANICS 59

IW2 do
Rotational kinetic energy KE =- If m is constant then force F=m-=m (mass x acceleration)
2 dt
where: I = moment of inertia of body Similarly: Torque T=- dUw) (rate of change of angular
m dt momentum)
Change of kinetic energy =- (u2 - u2)
2 dw
If I is constant T=I-=Ia
Potential energy PE = mgh dt
where: g=acceleration due to gravity (9.81 ms-2),
h = height above a datum. 2.1.10 Impact
kx2 The following deals with the impact of elastic and
Strain energy SE = Fx = - inelastic spheres, although it applies to bodies of any
2
shape.
where: x = deflection, k =stiffness. Consider two spheres rolling on a horizontal plane.
Velocities before impact are ul and u2 for spheres of
Conversion of potential energy to kinetic energy:
mass m, and m2. After impact their velocities are uI
mu2 and u2.
Wh=-
2
Coefficient of restitution
V2
Therefore v = J2gh or h = - difference in final velocities
29 e= - = _- (ul -v2)
Power difference in initial velocities (ul -uz)
Note: e=l for perfectly elastic spheres; e=O for
inelastic spheres.
Velocities after impact (velocities positive to right):
Rotational power P = torque x angular velocity
Te
=Tw=-
t
Also, if N = the number of revolutions per second
P=2nNT
where: 2nN =angular velocity w.

2. I .9 Impulse and momentum
Impulse. An impulsive force is one acting for a very
short time dt. Impulse is defined as the product of the
force and the time, Le. =Fat.
Momentum is the product of mass and velocity=mv
Change of momentum = mu - mu ml(u: -u:)+m2(u: -u;)
Angular momentum =Io Loss of kinetic energy due to impact = 2
Change of angular momentum = I(w, - w,) If e= 1, KE loss=O.

d (mu)
Force F = rate of change of momentum = -
dt