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=5.4g and radius
r=0.8cm . The meter stick has mass M=.16kg and length D=1m. (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=2.5m 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 v ? . 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.

(a) What is v ?com , the marble

Explanation / Answer

a) TE1 = m*g*h.........

TE2 = KEr + KEt = 0.5*(2/5)*m*r^2*w^2 + 0.5*m*Vcm^2

TE2 = 0.5*(2/5)*m*r^2*Vcm^2/r^2 + 0.5*m*Vcm^2

TE2 = 0.5*(7/5)*m*Vcm^2

TE2 = TE1

m*g*h = 0.5*(7/5)*m*V^2

Vcm = sqrt(2*(5/7)*g*h) = (2*(5/7)*9.8*2.5)^.5 = 5.92 m/s

b) L1 = m*Vcm*D = 0.0054*5.92*1 = 31.97e?3 kg m^2 /s

final MOI I2= (1/3)*M*D^2 + m*D^2 = (M/3 + m)D^2 = ((0.16/3)+0.0054)*1 = 58.73e?3 kg m^2

final angular momentum L2 = I2*W

according to law of conservation of angular momentum

L2 =L1

W = 31.97e?3/58.73e?3 = 544.36e?3 rad/s

d) KE = (M+m)*g*H

0.5*I2*W^2 = (M+m)*g*H

0.5*58.73e?3*(544.36e?3)^2 = (0.16+0.0054)*9.8*H

H = 5.37e?3 m

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