 #1
ac7597
 126
 6
 Homework Statement

Peti is the foreman of the Egyptian workers who have just finished building the great pyramid of Cheops.
L = 236 m
H = 140 m
M = 6 x 10^9 kg
It is late Friday afternoon, and everyone else has gone home for the day. The Grand Opening ceremony is scheduled for Monday.
The Vizier walks over to Peti and says, "I have good news and bad news. The good news is that it has the right size and shape. Well done."
"What's the bad news?" asks Peti.
"It's facing the wrong way. The face with the access tunnel is supposed to face NORTH, not South. The Big Guy is going to be very unhappy if you don't fix this by Monday morning."
Fortunately, Peti thought something like this might happen, so he placed a frictionless bearing under the very center of the pyramid. It will rotate freely  if someone can push it hard enough.
Peti runs over to one corner of the pyramid and starts to push, applying a force of F = 500 N.
What is the torque applied by Peti?
How long will it take Peti to rotate the pyramid by half a rotation?
How fast will Peti be moving at the maximum rotational speed?
What is the average power supplied over that time?
Will he finish in time?
 Relevant Equations
 torque=Force*radius*sin(theta)
torque=Force*radius*sin(theta)
center of mass x direction = ( 0(6 x 10^9 kg)+ (118m)(6 x 10^9 kg)+ (236m)(6 x 10^9 kg) )/(3(6 x 10^9 kg)) = 118m
center of mass y direction = ( 0(6 x 10^9 kg)+ (140m)(6 x 10^9 kg)+ (0)(6 x 10^9 kg) )/(3(6 x 10^9 kg)) = 46.7 m
radius = (118^2 + 46.7^2)^(1/2) = 126.9m
torque= 500N * 126.9m *sin(90deg)= 63.4*10^3 N*m
I am confused on how to proceed further
center of mass x direction = ( 0(6 x 10^9 kg)+ (118m)(6 x 10^9 kg)+ (236m)(6 x 10^9 kg) )/(3(6 x 10^9 kg)) = 118m
center of mass y direction = ( 0(6 x 10^9 kg)+ (140m)(6 x 10^9 kg)+ (0)(6 x 10^9 kg) )/(3(6 x 10^9 kg)) = 46.7 m
radius = (118^2 + 46.7^2)^(1/2) = 126.9m
torque= 500N * 126.9m *sin(90deg)= 63.4*10^3 N*m
I am confused on how to proceed further