posted by lost on .
A billiard ball that is initially at rest is given a sharp blow by a cue stick. The force is horizontal and is applied at a distance h = 2R/3 below the centerline, The speed of the ball just after the blow is v0 and the coefficient of kinetic friction between the ball and the billiard table is μk
find angular velocity just after blow, velocity once roles without slipping, and KE just after hit.
The linear momentum is p =m•vₒ.
An angular momentum is
L = p•(2•R/3) = m•vₒ•(2•R/3).
Moment of inertia of the sphere is
L = I• ωₒ= (2•m•R²/5)•ωₒ.
m•vₒ•(2•R/3) = (2•m•R²/5)•ωₒ.
ωₒ = 5•vₒ/3•R.
Since the force is applied below the center line, the spin is
backward, i.e., the ball will slow down,
ωₒ = - 5•vₒ/3•R.
M = F•R =μ•g•R is the torque of the friction force,
then an angular acceleration is
ε = M/I = μ•g•R/(2•m•R²/5) = 5• μ•g/2•R.
v =vₒ - a•t = vₒ- μ•g•t.
ω = ωₒ+ε•t= -(5•vₒ/3•R) +(5• μ•g/2•R)•t.
Since v =ω•R,
v = - (5•vₒ/3•R) •R +(5• μ•g/2•R) •R•t =
= - (5•vₒ/3)+ (5•μ•g/2) •t,
vₒ - μ•g•t = - (5•vₒ/3) + (5•μ•g/2) •t ,
vₒ + (5•vₒ/3) = μ•g•t + (5•μ•g/2) •t.
16•vₒ = 21•μ•g•t,
t = 16•vₒ/21•μ•g.
v = vₒ- μ•g•t = vₒ- μ•g•(16•vₒ/21•μ•g) =
= 5• 21 =0.238•vₒ.
KE1 = m• vₒ²/2 + I•ωₒ²/2 =
= m• vₒ²/2 + (2•m•R²/5)•( 5•vₒ/3•R)²/2 =
=19• m• vₒ²/18 =1.056• m• vₒ².
KE2 = m• v²/2 + I•ω²/2 =
= m• v²/2 + (2•m•R²/5)•v²/2•R² = 0.7•m•v²=
ΔKE = - W(fr) = 1.056• m• vₒ² - 0.0397•m•vₒ² =