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pietastesgood
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Homework Statement
1. A cylinder of radius R = 6.0 cm is on a rough horizontal surface. The coefficient of kinetic
friction between the cylinder and the surface is 0.30 and the rotational inertia for rotation
about the axis is given by MR2/2, where M is its mass. Initially it is not rotating but its
center of mass has a speed of 7.0m/s. After 2.0 s the speed of its center of mass and its angular
velocity about its center of mass, respectively, are:
A. 1.1m/s, 0
B. 1.1m/s, 19 rad/s
C. 1.1m/s, 98 rad/s
D. 1.1m/s, 200 rad/s
E. 5.9m/s, 98 rad/s
Answer is D.
2. As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s.
The circle is parallel to the xy plane and is centered on the z axis, 0.75m from the origin. The
component in the xy plane of the angular momentum around the origin has a magnitude of:
A. 0
B. 6.0kg · m2/s
C. 9.0kg · m2/s
D. 11 kg · m2/s
E. 14 kg · m2/s
Answer is C.
3. A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer
radius, are each free to rotate about a fixed axis through its center. Assume the hoop is
connected to the rotation axis by light spokes. With the objects starting from rest, identical
forces are simultaneously applied to the rims. Rank the objects according to their
angular momenta after a given time t, least to greatest.
A. all tie
B. disk, hoop, sphere
C. hoop, disk, sphere
D. hoop, sphere, disk
E. hoop, disk, sphere
Answer is A.
Homework Equations
1. Friction=Iα/R, thus α=2μg/r
1/2m(vi)^2 = 3/4m(vf)^2
ωf=ωi+αt
2. L=Iω
I=MR^2
3. L=Iω
I=MR^2
The Attempt at a Solution
1. I'm getting the angular velocity fine, but I just cannot for the life of me get the linear velocity correct. Using α=2μg/r, I get the angular acceleration as 98 rad/s^2. So the angular velocity at t=2 should be 196 rad/s, or about 200 rad/s. Since a=αr, the linear acceleration should be 98*0.06, which equals -5.88 m/s^2, since the angular acceleration causes the body to slow down. When plugging that into ωf=ωi+αt, I get a negative number for the final angular velocity, which would yield a negative linear velocity, which isn't right. When I try to do it in with the energy conservation formula I also get the wrong answer. What am I doing wrong in solving for the linear velocity?
2. Shouldn't this be zero? If you evaluate r x V using the right hand rule, the vector points up, so the angular momentum is solely in the z-plane, right?
3. Not quite sure how to approach this one. If I'd just guessed, I probably would have gone with A, simply because I would think that though the moments of inertia vary, a small moment of inertia would yield a larger angular velocity, so all three would end up with the same angular momentum, based off L=Iω. Is this reasoning correct?
Thanks in advance!