Fido 3:08 PM 97% Moodle 30° s about a 8. A uniform disk turns 4.40 s spindle. A
ID: 2304886 • Letter: F
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Fido 3:08 PM 97% Moodle 30° s about a 8. A uniform disk turns 4.40 s spindle. A nonrotating rod of the same mass as the disk and length equal to the disk's diameter is dropped onto the freely spinning disk where it sticks to the disk. They then both turn around the spindle, with their centers superimposed. What is the angular speed (in revs/s) of the combination? Rod: Im-, L-length of rod Disk: Im R-radius 9. A rigid rod of length 40.0 cm, and mass 0.500 kg is free to rotate in all the direction about a pivot point at one end of the rod located at the origin. The rod is initially resting along the positive x-axis. In addition to gravity, a constant force is acting on the rod at point 30.0 cm away from the pivot point. (Recall that the force of gravity is along the -z axis). a) Find the net torque acting on this rod. b) What is the rotational moment of inertia, I, of the rod about the pivot point? (show your work) (note: Im) e) Find the angular acceleration of this rod. . A pulley and mass system is setup as shown below. The pulley is not massless, but you may assume there is no slipping of the string along the pulley and that the string is massless. The system starts from rest and the hanging mass is seen to move upwards. There is a tension in the left portion of the string, and a tension in the right portion of the string. Is the magnitude of larger or smaller than the magnitude of Explain your answer in full detail, state any assumptions you have made (You are being marked on your ability to explain the physics of what is happening) (5 marks)Explanation / Answer
8. given w = 4.4 rev/s = 8.8*pi rad/s
uniform disc, mass m, diameter = d
non rotating rod of mass m, lenggth = d
let final angular speed be w
then from conservation of angular momentum
0.5md^2*w/4 = (0.5md^2/4 + md^2/12)w'
3*w = (5)w'
w' = 3w/5 = 2.64 rev/s = 5.28*pi rad/s
9. l = 40 cm
m = 0.5 kg
pivot point is at one end of the rod
force F is acting upwards at r = 30 cm
hence from torque balance
F*30 = mg*40/2
a. net torque on this rod = 0
because this rod is in equilibrium
b. rotational moment of inertia about the pivot point = ml^2/3 = 0.02666666666666 kgm^2
c. angular acceleration = alpha = 0
as torque on the rod is 0
10. as the hanging mass is moving upwards, hecne
if the hanging mass is on the right, the tension on the left string is more than the right string, and that is why the hanging mass is moving upwards
for the hanging mass to the left, then the right string has more tension thatn the left strgin
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