A couple problems with angular momentum

In summary, the conversation includes a request for help with a physics problem involving a rolling ring on an incline plane, a uniform solid disk rotating about a fixed axis, and a solid sphere rolling without slipping on a track. The conversation also discusses calculations and potential errors in the solutions.
  • #1
Nanabit
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I know a couple of these I already asked, but I'm really close and missing something. Could someone help??

1) 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 = (1/2)*M*(R1^2+R2^2) = .01265 kg m^2.
v = rw
2.75 m/s = .08mw
w=34.375 rad/sec

w = omega.

(1/2)mv^2+(1/2)Iw^2 = mgh (I'm setting the initial mgh to zero.)
(1/2)(2.53kg)(2.75m/s)^2 + (1/2)(.01265 kg m^2)(34.38)^2 = (2.53kg)(9.8m/s^2)(sin 36.4)x
x = 1.158 m

initial distance up at the x = 2 mark on the plane is 1.19m.

1.158 + 1.19 = 2.35 m

What's wrong??

2) A uniform solid disk of mass 2.98 kg and radius 0.200 m rotates about a fixed axis perpendicular to its face. Omega = 5.95 rad/sec.
What is the angular momentum when the axis of rotation passes through a point midway between the center and the rim?

I = I(i)+mr^2
I = .355kgm^2+(2.98kg)(.1^2)
I = 2.30 kg m^2 / sec

What's wrong?

3) A solid sphere of mass m and radius r rolls without slipping along the track. 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.

What are the force components on the sphere at the point P if h = 3R?

I got Fx being -20/7mg, and I thought Fy would be -mg but it isn't.

THanks.
 
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  • #2
1. At first glance, your solution looks good.

initial distance up at the x = 2 mark on the plane is 1.19m.

Is it possible that when they say 2 m up the plane, they mean along the plane (making your final answer 2 + 1.158)?

2. Be careful here. I think you meant to write:

Io = .5*M*R^2 = .0596 kgm^2

I = .5*M*R^2 + Mh^2 = .0596 + 2.98*(.1)^2

andular momentum L = Iω = I*5.95rad/s (=.53 kg m^2/s)

3. Where is point P? Somewhere on the loop? If you get a chance, could you show where your answer came from too?
 
  • #3


1) In the first problem, it seems like you may have made a calculation error when finding the initial distance up the plane. You calculated it to be 1.19m, but it should be 1.16m (rounded to two significant figures). This could have affected your final answer of 2.35m. Also, you should not set the initial mgh to zero, as this is the potential energy at the starting point and should be included in the total energy equation.

2) In the second problem, you have correctly calculated the moment of inertia, but you have not used it in the calculation of angular momentum. The correct formula for angular momentum is L = Iω, where I is the moment of inertia and ω is the angular velocity. So the angular momentum in this case would be 2.30 kg m^2/s, not just 2.30 kg m^2/s^2.

3) In the third problem, it seems like you may have forgotten to include the normal force in your calculation for Fx. The normal force is responsible for the horizontal force component, while gravity is responsible for the vertical component. So the correct formula for Fx would be Fx = -20/7mg + N, where N is the normal force. Also, your calculation for Fy is correct, as gravity is the only force acting in the vertical direction.
 

1. What is angular momentum?

Angular momentum is a physical quantity that describes the rotational motion of an object around an axis. It is a vector quantity, meaning it has both magnitude and direction, and is conserved in a closed system.

2. How is angular momentum calculated?

The angular momentum of an object can be calculated by multiplying its moment of inertia (a measure of its resistance to rotational motion) by its angular velocity (how fast it rotates around an axis) and by the sine of the angle between the two vectors.

3. What are the units of angular momentum?

The SI unit of angular momentum is kilogram meters squared per second (kg*m^2/s), but it can also be expressed in other units such as Joule seconds (J*s) or erg seconds (erg*s).

4. How does angular momentum conservation apply to everyday life?

Angular momentum conservation can be observed in many everyday situations, such as when a spinning top stays upright due to its angular momentum, or when a figure skater spins faster by pulling their arms in closer to their body, conserving their angular momentum.

5. What are some common problems with angular momentum?

Some common problems with angular momentum include confusion with linear momentum (which describes the motion of an object in a straight line), difficulty in visualizing rotational motion, and the complexity of calculations involving multiple axes of rotation.

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