QUESTION 13
A chimney (length 13.4 m, mass 648 kg] cracks at the base and topples. Assume:
- the chimney behaves like a thin rod, and does not break apart as it falls.
- only gravity (no friction) acts on the chimney as it falls.
- the bottom of the chimney pivots, but does not move.
Find the linear speed of the center of mass of the chimney, in m/s, just as it hits the ground.
QUESTION 15
A 75.1 kg cart travels at constant speed, and its four wheels are rolling without slipping. The wheels are disks (radius 0.771 m, mass 8.82 kg), and they turn at constant ω = 50.4 rad/s. Find the total kinetic energy of the cart.
NOTE: first, you must find the speed of the cart!
Question 13:
given
L = 13.4 m
m = 648 kg
let v is the speed of center of mass and w is the angular speed of
the chimney when it hits the ground.
moment of inefrtia of the chimney about bottom, I = m*L^2/3
Apply conservation of energy
final kinetic energy = initial potential energy
(1/2)*I*w^2 = m*g*h
(1/2)*(m*L^2/3)*w^2 = m*g*(L/2)
w^2*L/3 = g
w = sqrt(3*g/L)
= sqrt(3*9.8/13.4)
= 1.481 rad/s
now use, v = r*w
= (L/2)*w
= (13.4/2)*1.481
= 9.92 m/s <<<<<<<<<---------------------Answer
Q 15)
given
M = 75.1 kg
r = 0.771 m
m = 8.82 kg
w = 50.4 rad/s
speed of the cart, v = r*w
= 0.771*50.4
= 38.86 m/s
total kinetic energy of the cart = (1/2)*M*v^2 + 4*(1/2)*I*w^2
= (1/2)*M*v^2 + 4(1/2)*0.5*m*r^2*w^2
= (1/2)*75.1*38.86^2 + 4(1/2)*8.82*0.771^2*50.4^2
= 83340 J <<<<<<<<-----------------Answer
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