Question

A 5 lb. pendulum (1) is released from rest when the angle is 0 ̊. A 2-lb crate (2) is at originally at rest on a rough surface µ = 0.2. The pendulum strikes the crate at an angle 90 ̊ with a coefficient of restitution of 0.8. The crate slides on a frictionless curve surface until it became to rest by a linear spring (D). If the cord has L= 2.25 ft long;

1.Determine the force of tension of the cord at impact point (B); [ lb ].

2. Determine de velocity of the crate after the impact ft/s

3. Determine: the velocity of the crate at the beginning of the smooth surface @ (c). [ ft/s]

3-ft 2 u=0.2 с = 0 B D 2.2 ft NOK

Answer 1

0=0 0=90° 1. Using conservation of Energy, L 1 L Energy 8-98 L Energy 0:0° mg 2 La mua v=Sagl L=2.25 ft V2x32.2x2.25 = 12.03 ft's Tension in Cord, T= mg + + Fc • 5+ 5x12.032 2.25 (Fc = Centrifugal force ) m v? mv2 Ing y & Fc meslb T: 326.602 lb ui, vi pendulum velocities before and after impact U2, V₂ Coate velocities before and after impact U, = 12,03 ft/s Coefficient of Restitution, e: V2-vi u,-42 U2=0 V2-vi 0.8 = 12.03 V2-V, = 9.624 Using Conservation of lineal momentum, m, ut malla = mult mavz 5*12.03 + 0 svit 2V2 5V,tav2 = 60.15 Solving O and V, = 5,843 ft/s V2 = 15.467 ft/s velocity of crate after impact: Ya= 15.469 ft/s

3. N friction, f= MN = ung 0.4 lbf 0.2*2 : f ing Fnet - f = - img - - ug -0,2% 32,2 -6.44 ft/s2 m m Consider V-V: 225 (s: 34t, v; = 15.467 ft/s) V$-15.467" = - 2x6.44X3 4:14.163-14 Velocity of Crate at beginning of smooth Surface, Vc = 14.163 ft/s