1. 1 points Tipler6 10.P064 My Notes Ask Your Teac The figure below shows a thin

Question

1. 1 points Tipler6 10.P064 My Notes Ask Your Teac The figure below shows a thin, uniform bar of length D 1.12 m and mass M 0.79 kg pivoted at the top. The rod, which is initially at rest, is struck by a particle whose mass is m = 0.30 kg at a point x0.80d below the plvot. Assume that the particle sticks to the rod. If the maximum angle between the rod and the vertical following the collision is 60, find the speed of the particle before impact. m/s eBook Subemit Answerl Save Progress Practice Another Version

Explanation / Answer

after the particel strucks to the rod

center of mass Xcm = (M*d/2) + (m*x)/(M+m) = ((0.79*1.12/2)+(0.3*0.8*1.12))/(0.79+0.3) = 0.652 from pivot

potential energy of the system at 60 degrees PE = (M+m)*g*Xcm*(1-cos60)

PE = (0.79+0.3)*9.8*0.652*(1-cos60) = 3.48 J

from energy conservation

kinetic energy after collision = PE

(1/2)*I*w^2 = PE

I = (1/3)*M*d^2 + m*x^2

I = (1/3)*0.79*1.12^2 + 0.3*(0.8*1.12)^2 = 0.571 kg m^2

(1/2)*0.571*w^2 = 3.48

angular speed w (omega) = 3.5 rad/s

angular momentum before collision Li = m*v*x

angular momentum after collision Lf = I*w

from conservation angular momentum

total angular momentum remains constant

Li = Lf

m*v*x = I*w

0.3*v*0.8*1.12 = 0.571*3.5

speed of particel v = 7.434 m/s <<<-------------ANSWER

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.