posted by RON on .
A 20lb ball is 1 ft. in diameter and has a angular velocity of 300 rpm when it is 2 ft. away from the plate. The coefficient of rolling friction between the surface and the ball is μs = 0.3. It strikes the 1-ft. high, 10-lb plate that is held in place by an unstretched spring of stiffness k = 400 lb/ft. Take e = 0.8 between the ball and the plate and assume that the plate slides smoothly.
Find the maximum compression imparted to the spring.
I used to work with SI units:
m1 = 20 lb = 9.07 kg,
m2 = 10 lb = 4.536 kg,
R = 1ft = 0.305 m,
s =2 ft = 0.61 m.
k =400 lb/ft = 5837 N/m,
μ = 0.3,
coefficient of restitution e = 0.8.
For the ball:
n1 = 300 rpm = 5 rev/s =>
ω1 =2• π•n(b1) = 10•π =>
v1 = ω(b1) •R = 10•π•0.305 =9.58 m/s.
The kinetic energy of the rolling ball at its initial position is
KE1 =m•v1^2/2 +I •ω1^2/2 = m•v1^2/2 +(2•m•R^2/5) • (v1^2/2•r^2) = 0.7•m1•v1^2 =0.7•9.07•(9.58)^2 = 583 J.
KE1 – KE2 =W(friction)
W(friction) = μ•m•g•s =0.3 •9.07•9.8•0.61 = 16.27 J.
KE2 = KE1 - W(friction) = 583 - 16.27 = 566.73 J.
v2 = sqrt(2•KE2/m1) = sqrt(2•566.73/9.07) = 11.18 m/s.
Law of conservation of linear momentum:
m1•v2 = m1•v3 +m2•u
According to the definition the coefficient of restitution is
e = (u – v3)/v2,
e•v2 = u – v3
v3 = u - e•v2.
m1•v2 =m1• (u - e•v2) + m2•u = m1• u - m1• e•v2 + m2•u.
u• (m1+m2) = m1•v2• (e + 1 )
u = m1•v2(e + 1 )/ (m1+m2) =
9.07•11.18•1.8/(9.07 + 4.536) = 13.42 m/s.
k•x^2/2 = m2•u^2/2,
x =u•sqrt(m2/k) = 13.42•sqrt(4.536/5837) = 0.37 cm ≈ 1.2 ft.
Maybe I’ve mistaken in calculations (especially due to the units transformation), but the general form is correct. If you know the answer, post it