Posted by **RON** on Monday, April 30, 2012 at 7:42am.

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.

- Physics -
**Elena**, Monday, April 30, 2012 at 1:00pm
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.

KE2 =m1•v2^2/2,

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

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