of plywood of mass 8.00 kg and four wheels, each 20.0 cm in diameter an with a m
ID: 1789667 • Letter: O
Question
of plywood of mass 8.00 kg and four wheels, each 20.0 cm in diameter an with a mass of 2.00 kg. It is released from the top of a 15.0° incline that is 30.0 m long. Find the speed at the bottom. Assume that the wheels roll along the incline without slipping and that friction between the wheels an their axles can be neglected. 10.76 A CD has a mass of 15.0 g, an inner diameter of 1.50 cm, and an oute diameter of 11.9 cm. Suppose you toss it, causing it to spin at a rate of 4.30 revolutions per second. Determine the moment of inertia of the CD, approximating its density a) as uniform. b) If your fingers were in contact with the CD for 0.250 revolutions while i was acquiring its angular velocity and applied a constant torque to it, what was the magnitude of that torque? 10.77 A sheet of plywood 1.30 cm thick is used to make a cabinet door 55.0 cm wide by 79.0 cm tall, with hinges mounted on the vertical edge. A small 150.-g handle is mounted 45.0 cm from the lower hinge at the same height as that hinge. If the density of the plywood is 550. kg/m, what is the moment of inertia of the door about the hinges? Neglect the contribution of hinge components to the moment of inertia. 10.78 A machine part is made from a uniform solid disk of radius R and mass M. A hole of radius R/2 is drilled into the disk, with the center of the hole at a distance R/2 from the center of the disk (the diameter of the hole spans from the center of the disk to its outer edge). What is the moment of inertia of this machine part about the center of the disk in terms of R and M? 10.79 A space station is to provide artificial gravity to support long-term habitation by astronauts and cosmonauts. It is designed as a large wheel, with all the compartments in the rim, which is to rotate at a speed that will provide an acceleration similar to that of terrestrial gravity for the astronauts (their feetExplanation / Answer
The mass of the disk = M
The radius of the disk = R
hence the density of the material = M/piR^2
The net moment of Inertia will be equal to the moment of Inertia of the solid disk minus the moment of Inertia of the hole.
Area of the hole = pi R^2/4
Parallel theorem of moment of inertia
I = I_c + M d^2 where d is the distance between parallel axis.
I = M R^2/2 - [(M/pi R^2)*piR^2/4] * (R^2/8 + R^2/4)
I = MR^2/2 - (M/4)*3R^2/8 = MR^2/2 - 3MR^2/32
I = 13 MR^2/32
I = (13/32) MR^2
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