A glass marble is allowed to roll down a curved ramp so that it collides with th

Question

A glass marble is allowed to roll down a curved ramp so that it collides with the end of a wooden meter stick that is suspended from a frictionless pivot at point P, as shown below. The marble has mass m = 5.40 g and radius r = 0.80 cm. The meter stick has mass M = 0.160 kg and length D = 1.0 m. (The figure is NOT drawn to scale.) At first, the marble is held at rest on the ramp with its center of mass a height h = 2.50 m above the lower end of the meter stick. After it is released from rest, the marble rolls smoothly down the ramp and onto a horizontal surface where it rolls with constant horizontal center of mass velocity vcom. When the marble collides with the meter stick, a patch of super-sticky glue causes the marble to immediately adhere to the end of the meter stick. (Note that this figure is NOT drawn to scale.) For all parts of this problem, you must work the problem symbolically to find a final symbolic answer before using the provided numbers to find a final numerical answer. What is , the marble's COM velocity as it rolls along the horizontal surface? Report your answer in unit vector notation using the coordinates provided. What is , the angular momentum of the marble (about point P) just before the collision? (You only need to consider the translational motion of the marble's COM. You should investigate for yourself why this is true.) Report your answer in unit vector notation using the coordinates provided. What is , the angular velocity of the marble-stick combination immediately after the collision? Report your answer in unit vector notation using the coordinates provided. Through what height H does the COM of the marble-stick combination move before it comes to rest?

Explanation / Answer

a) m*g*h = 0.5*m*v^2 + 0.5*I*w^2

m*g*h = 0.5*m*v^2 + 0.5*(2/5*m*r^2)*w^2

m*g*h = 0.5*m*v^2 + 0.2*m*v^2

m*g*h = 0.7*m*v^2

==> vcm = sqrt(g*h/0.7)

   = sqrt(9.8*2.5/0.7)
   = 5.916 m/s <<<<<<<-----Answer
b) L1 = m*v*D
      = 5.4*10^-3*5.916*1
      = 0.03195 kg.m^2/s <<<<<<--Answer

c) I2 = M*D^2/3 + (2/5)*m*r^2 + m*D^2

       = 0.16*1^2/3 + 0.4*0.0054*0.008^2 + 0.0054*1^2

       = 0.05873 kg.m^2

L2 = I2*w

here, L2 = L1

I2*w = L1

w = L1/I2
= 0.03195/0.05873

= 0.5444 rad/s <<<<<<<-Answer

d) kinetic after the collsion, KE = 0.5*I2*w^2
= 0.5*0.05783*0.544^2

= 0.00856 J

(M+m)*g*H = 0.00856

H = 0.00856/(M+m)*g

= 0.00856/(0.16+0.0054)*9.8

= 5.27 mm

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