Rotational Motion Problems: Tilting Pole, a Yo-Yo, and a rolling hoop

[random26]

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 :)

 

[kuruman]

Problem 1
How is radius of a pole, which is pivoted and rotates about one end, related to its length of 2.30 m?
Problem 2
By what vertical distance is a yo-yo displaced when it drops from one end to the other of a 1-m long string?
Problem 3
You need to use the kinematics equations involving time for the second part.

 

[haruspex]

its lower end does not slip

Is there a difficulty here?
When the pole length 2r is at angle θ to the ground, KE gives:
$\frac 12\frac 43mr^2\omega^2=mgr(1-\cos(\theta))$
Centripetal acceleration gives:
$mr\omega^2=mg\cos(\theta)-N\cos(\theta)-F\sin(\theta)$
where N is the normal force and F the frictional force:
$|F|\leq |N\mu_s\sin(\theta)|$
When $\cos(\theta)=\frac 35$ those equations become:
$mr^2\omega^2=mgr\frac 35$
$mr\omega^2=mg\frac 35-N\frac 35-F\frac 45$
Whence
$N\frac 35+F\frac 45=0$.
So it seems that even without slipping the pole will lose contact with the ground.

the radius of the pole

Not sure what you mean by that, the radius of rotation of the mass centre of the pole or its radius as a solid cylinder.

Would the height be zero?

You can choose the zero height wherever you like. All that will matter is the change in height.

When the yo-yo is descending at speed v, how fast is it rotating? What is its KE?