Bowling Ball Angular momentum problems

[Nanabit]

There are only 2 problems I am absolutely clueless on; if someone could help me get started on those 2 problems and help me figure out where I went wrong on the others, that would be great.

1. A bowling ball has a mass of 5.00 kg, a moment of inertia of 1.60 10-2 kg · m2, and a radius of 0.100 m. If it rolls down the lane without slipping at a linear speed of 5.00 m/s, what is its total energy?

I figured, through adding rotational KE for a sphere to the normal formula for KE, that the total KE formula would be (7/10)mv^2. I got 87.5 J, which isn't right. Should I not be using this formula?

2. A ring of mass 2.53 kg, inner radius 6.00 cm, and outer radius 8.00 cm is rolling (without slipping) up an incline plane which makes an angle of = 36.4°. At the moment the ring is at position x = 2.00 m up the plane, its speed is 2.75 m/s. The ring continues up the plane for some additional distance, and then rolls back down. It does not roll off the top end. How far up the plane does it go?

I can figure out the moment of inertia for the ring, no problem. I got .01265 kgm^2 for that. Then I thought I could figure out the final velocity by using conservation of momentum. (2.53kg)(9.80m/s^2)(2.00m))+(1/2)(2.53kg)(2.75 m/s)^2 = (1/2)(2.53kg)(vf^2). I got the final velocity to be 5.55 m/s. I don't know where to go from there. I thought if I could find acceleration I could use kinematics to figure out the final x distance, but I don't know how to find acceleration.

3. A force of F = 3.00 i + 2.00 j N is applied to an object that is pivoted about a fixed axis aligned along the z coordinate axis.
(a) If the force is applied at the point r = (4.00 i + 7.00 j + 0 k) m, find the magnitude of the net torque about the z axis. N · m
(b) What is the direction of the torque vector?

I have no problem with part a, I found that to be 13Nm. I just don't know how to tell the direction of the torque vector.

4. A uniform solid disk of mass 2.98 kg and radius 0.200 m rotates about a fixed axis perpendicular to its face.
(a) If the angular speed is 5.95 rad/s, calculate the angular momentum of the disk when the axis of rotation passes through its center of mass.
(b) What is the angular momentum when the axis of rotation passes through a point midway between the center and the rim?

Once again, no problem with part a. Part b I thought I could just divide the radius by 2 and multiply that by the angular speed, but it isn't right.

5. The hour and minute hands of Big Ben, the famous Parliament Building tower clock in London, are 2.60 m and 4.60 m long and have masses of 59.0 kg and 99 kg, respectively. Calculate the total angular momentum of these hands about the center point. Treat the hands as long thin rods.

I'm lost.

6. A playground merry-go-round of radius R = 1.60 m has a moment of inertia I = 235 kg · m2 and is rotating at 12.0 rev/min about a frictionless vertical axle. Facing the axle, a 24.0 kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round?

This one makes perfect sense to me, but it's not working out. I figured first I had to figure out the initial mass. I = (1/2)mr^2. 235 kg = m (1.60m)^2. mass = 183.6 kg. Then add for the final moment of inertia the weight of the child. I = (1/2)mr^2. I = (1/2)(207.6kg)(1.60m)^2. I = 265.7Kgm^2. Then use the formula Iiwi = Ifwf. (235kgm^2)(1.26rad/sec)=(265.7kgm^2)(wf). I got 1.11 rad/sec to be the answer, but the answer has to be in rev/min, so I got 10.6 rev/min and have no idea why it is wrong.

7. A solid sphere of mass m and radius r rolls without slipping along the track shown in Figure P11.51. The sphere starts from rest with the lowest point of the sphere at height h above the bottom of the loop of radius R, which is much larger than r. a) What is the minimum value that h can have if the sphere is to complete the loop? (Use R for R, m for m, r for r, and g for gravity, as necessary.)
(b) What are the force components on the sphere at the point P if h = 3R?
Fx = N
Fy = N

I'm lost.

Thanks again for any help.

 

[jamesrc]

1. Moment of inertia of a sphere about its center is (2/5)mR^2 so when you add rotational to kinetic energy, you should get (9/10)mv^2

2. Try using conservation of energy here; find the potential energy when the kinetic energy = 0 (when the ring has stopped). So K_rot(x=2) + K_tran(x=2) + U(x=2) = U(x=?)

where (x=?) stands for the function evaluated at x = ?. Potential energy due to gravity is just mgh as usual.

3. Two ways:
a.) use the right hand rule: Draw your picture looking at the top of it (so the z-axis is pointing out of the page at you, y-axis is pointing up, and x-axis is pointing to the right). Draw your force vector at the distance given and decide which way the force will make the object rotate. For this right-handed system, counterclockwise is positive (+z) and clockwise is negative (-z). You can see this by curling the fingers on your right hand from the position vector to the force vector. The direction your thumb is pointing is the direction of the force. (Incidentally, I don't think it's a problem if you answer a question like that as "clockwise" or "counter-clockwise"
b.) you can do it analytically. torque is the vector cross product of moment arm and force: τ = r x F
(bold indicates vectors). I won't go into that since I don't know if you know it or how well you know it, but it's there to look up if you're interested.

4. For the second part of this one, use the parallel axis theorem. This allows you to calculate the moment of inertia about a point other than the point you already know it at. So I_pa = I_o + MR^2 where I_pa is what you're solving for, I_o is the moment of inertia about its center (which you already found in the first part), and R is the distance between the old axis and the new axis. For the angular speed, I would guess that they want you to use the same ω (you shouldn't have to do anything to it; it's just a given).

5. Can you calculate the moment of inertia for the rods? If so, you're almost done. You get the speed using what you know about a clock (assume constant angular velocities): The angular speed for the minute hand is 2π radians in 1 hour. The angular speed for the hour hand is 2π radians in 12 hours.

 

[M98Ranger]

1. For this problem, you can use the formula for total energy which is the sum of kinetic energy and potential energy. In this case, the bowling ball has both translational and rotational kinetic energy. So the formula should be: E = 1/2mv^2 + 1/2Iω^2, where I is the moment of inertia and ω is the angular velocity. In this case, the ball is not rotating, so ω = v/r. Plugging in the values, we get E = 1/2(5.00 kg)(5.00 m/s)^2 + 1/2(1.60 x 10^-2 kg m^2)(5.00 m/s)^2 = 12.5 J + 0.4 J = 12.9 J.

2. To find the distance the ring goes up the incline, you can use conservation of energy. The initial energy at the bottom of the incline is equal to the final energy at the top of the incline. The initial energy includes both translational and rotational kinetic energy, while the final energy only has potential energy. So the equation will be: 1/2Iω^2 + 1/2mv^2 = mgh, where h is the height the ring goes up the incline. You already calculated the moment of inertia, so you just need to plug in the values for ω and v and solve for h.

3. The direction of the torque vector can be found using the right-hand rule. If you point your fingers in the direction of the force, and curl your fingers towards the direction of rotation, your thumb will point in the direction of the torque vector. In this case, the torque vector will be in the negative z-direction.

4. For part b, you need to use the parallel axis theorem. The moment of inertia for a disk rotating around an axis through its center is 1/2mr^2. So, for a disk rotating around an axis through a point halfway between the center and the rim, the moment of inertia will be 1/2mr^2 + 1/2mr^2 = mr^2. So you can use the same formula as part a, but plug in the new moment of inertia.

5. To find the total angular momentum of the hands, you can treat each hand as a thin rod rotating around its center of