- #1
random26
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Okay, 3 problems I can't seem to get the right answer for..
1. A 2.30-m-long pole is balanced vertically on its tip. It starts to fall and its lower end does not slip. What will be the speed of the upper end of the pole just before it hits the ground? [Hint: Use conservation of energy.]
GPE = TKE + RKE
mgh = .5m(v^2) + (.5)(I)(w^2)
so: mgh = .5m(v^2) + (.5)[(1/3)(m)(L^2)]w^2
My problem is that I don't understand how to solve for the angular velocity [w] without the radius of the pole. The answer should be 8.22m/s. Suggestions?
2. A yo-yo is made of two solid cylindrical disks, each of mass .050kg and diameter .075m, joined by a (concentric) thin solid cylindrical hub of mass .0050kg and diameter .01m. Use conservation of energy to calculate the linear speed of the yo-yo when it reaches the end of its 1.0-m-long string, if it is released from rest. (b) What fraction of its kinetic energy is rotational?
I honestly don't even know where to begin with this problem. I guess it would have GPE at the top, and KE and GPE at the bottom, right? I don't understand how to know the height of the yo-yo at the top though for GPE = mgh. Would the height be zero? I definitely need help just getting started on this problem. The answer should be .84 m/s and 96%.
3. A hollow cylinder (hoop) is rolling on a horizontal surface at speed v=3.3m/s when it reaches a 15 degree incline. (a) How far up the incline will it go? (b) How long will it be on the incline before it arrives back at the bottom?
So it has TKE and RKE at the beginning and GPE once it's on the incline so..
.5(m)(v^2) + (.5)(m)(v^2) = mgh
therefore: (.5)(3.3^2) + (.5)(3.3^2) = (9.8)(h)
h ends up being 1.11 and when you divide it to get the hypotenuse, the answer is 4.3m up the ramp. So far so good. But I don't understand how you can relate this answer to the time you need to find in part b. Is there an equation for this I can use? The answer's supposed to be 5.2s, but I have no idea how to find that.
Thanks :)
1. A 2.30-m-long pole is balanced vertically on its tip. It starts to fall and its lower end does not slip. What will be the speed of the upper end of the pole just before it hits the ground? [Hint: Use conservation of energy.]
GPE = TKE + RKE
mgh = .5m(v^2) + (.5)(I)(w^2)
so: mgh = .5m(v^2) + (.5)[(1/3)(m)(L^2)]w^2
My problem is that I don't understand how to solve for the angular velocity [w] without the radius of the pole. The answer should be 8.22m/s. Suggestions?
2. A yo-yo is made of two solid cylindrical disks, each of mass .050kg and diameter .075m, joined by a (concentric) thin solid cylindrical hub of mass .0050kg and diameter .01m. Use conservation of energy to calculate the linear speed of the yo-yo when it reaches the end of its 1.0-m-long string, if it is released from rest. (b) What fraction of its kinetic energy is rotational?
I honestly don't even know where to begin with this problem. I guess it would have GPE at the top, and KE and GPE at the bottom, right? I don't understand how to know the height of the yo-yo at the top though for GPE = mgh. Would the height be zero? I definitely need help just getting started on this problem. The answer should be .84 m/s and 96%.
3. A hollow cylinder (hoop) is rolling on a horizontal surface at speed v=3.3m/s when it reaches a 15 degree incline. (a) How far up the incline will it go? (b) How long will it be on the incline before it arrives back at the bottom?
So it has TKE and RKE at the beginning and GPE once it's on the incline so..
.5(m)(v^2) + (.5)(m)(v^2) = mgh
therefore: (.5)(3.3^2) + (.5)(3.3^2) = (9.8)(h)
h ends up being 1.11 and when you divide it to get the hypotenuse, the answer is 4.3m up the ramp. So far so good. But I don't understand how you can relate this answer to the time you need to find in part b. Is there an equation for this I can use? The answer's supposed to be 5.2s, but I have no idea how to find that.
Thanks :)