Resolving Forces and Calculating Rotational Inertia of Assemblies

[BnJ]

1. Homework Statement :
In the figure, a stationary block explodes into two pieces L and R that slide across a frictionless floor and then into regions with friction, where they stop. Piece L, with a mass of 2.2 kg, encounters a coefficient of kinetic friction µL = 0.40 and slides to a stop in distance dL = 0.15 m. Piece R encounters a coefficient of kinetic friction µR = 0.50 and slides to a stop in distance dR = 0.30 m. What was the mass of the original block?
? kg

Split the problem into two parts: explosion and then slowing. The explosion involves internal forces and cannot change the momentum. Did you write an equation for the conservation of momentum, using symbols where you don't have a value? To get the speeds of the two pieces, you need to recall how kinetic friction can slow an object until it slides to a stop. Using the sliding distances should give you the speeds.
Section 9-7

2. Homework Equations
I have tried this problem with two different approaches, Ffric=MkN F=ma p=mv

and Ml+rV=0= MlVl-MrVr=0
using Ke=1/2mv^2 to find the mass

3. The Attempt at a Solution
using both equations i get the answer 1.39 but this is not correct?



1. Homework Statement :
A small disk of radius r = 2.00 cm has been glued to the edge of a larger disk of radius R = 4.00 cm so that the disks lie in the same plane. The disks can be rotated around a perpendicular axis through point O at the center of the larger disk. The disks both have a uniform density (mass per unit volume) of 1.10 103 kg/m3 and a uniform thickness of 4.50 mm. What is the rotational inertia of the two-disk assembly about the rotation axis through O?
?kg·m2

Density is the ratio of mass to volume. The rotational inertia of a disk about its central axis is given in Table 10-2. The rotational inertia about an axis shifted from the central axis is given by the parallel-axis theorem.



2. Homework Equations
I=1/2Mr^2 for a cylinder
and parallel-axis theorem Icom+ML^2


3. The Attempt at a Solution
I get hung up knowing that to do with the densities and uniform thickness.
I don't know how to deal with them since they are glued together.
I have tried to find both I values and add them together then use parallel axis but this is not right

1. Homework Statement :
In the figure below, a 48.0 kg uniform square sign, of edge L = 2.00 m, is hung from a horizontal rod of length dh = 3.00 m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance dv = 4.00 m above the point where the rod is hinged to the wall.
(a) What is the tension in the cable?
N

(b) What are the magnitude and direction of the horizontal component of the force on the rod from the wall? (Include the sign. Take the positive direction to be to the right.)
N

(c) What are the magnitude and direction of the vertical component of this force? (Include the sign. Take the positive direction to be upward.)
N

Did you write A balance-of-torques equation, using the hinge as the rotation axis for calculating torques? Did you apply the force due to the sign's weight midway between the sign's attachment points? Do you recall how to calculate a torque given a force's magnitude and angle? After you get the tension, did you apply A balance-of-forces equation horizontally and vertically?
Section 12-5

2. Homework Equations



3. The Attempt at a Solution
I have no clue where to begin...sorry


1. Homework Statement :
A uniform beam is 5.0 m long and has a mass of 56 kg. In the figure below, the beam is supported in a horizontal position by a hinge and a cable, with angle θ = 50°. In unit-vector notation, what is the force on the beam from the hinge?
hinge = ( N)i hat + ( N)j hat

Did you write A balance-of-torques equation, using the hinge as the rotation axis about which to calculate torques? (Did you recall how to calculate a torque from a force's magnitude and direction?) Did you write A balance-of-forces equation for vertical force components? For horizontal force components?

2. Homework Equations



3. The Attempt at a Solution
I have no clue where to begin...sorry

 

[dynamicsolo]

OK, I'm going to start off by telling you that you will do yourself a huge favor in the future by only posting one problem per thread. I suspect no one has taken this on after 28 hours because of the amount of time it would take to deal with all this...

...What was the mass of the original block?

3. The Attempt at a Solution
using both equations i get the answer 1.39 but this is not correct?

I'll give you another tip, as this is a pit everybody falls into at one time or another -- especially on timed exams. It is a natural, but often regrettable, human impulse to get through a particularly difficult intermediate calculation, arrive at an answer, say "Ah! Finished!", circle the result and move on. I warn all the students I work with that, when you think you're done with a problem, always go back and read the question again.

I completely agree with you that the mass of fragment R is 1.39 kg. What does the question ask you for?

...What is the rotational inertia of the two-disk assembly about the rotation axis through O?
?kg·m2
[/U][/B]


...
3. The Attempt at a Solution
I get hung up knowing that to do with the densities and uniform thickness.
I don't know how to deal with them since they are glued together.
I have tried to find both I values and add them together then use parallel axis but this is not right

You will need the mass of each disk, so how do you compute those?

I see a problem with your last statement. The big disk is the one whose center is on the rotation axis, so its center-of-mass rotational inertial is just (1/2)M(R^2). You need to add the rotational inertia of the small disk, which is the sum of its center-of-mass rotational inertia, (1/2)m(r^2), plus the term m(L^2) from the parallel-axis theorem. What is the distance of the center of mass of the small disk from the rotation axis?

The total rotational inertia of the assembly relative to O is then the sum of the two individual disks' rotational inertias relative to O. (You don't add the CM inertias of both and then apply the parallel-axis theorem to the set.)

In the figure below, a 48.0 kg uniform square sign, of edge L = 2.00 m, is hung from a horizontal rod of length dh = 3.00 m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance dv = 4.00 m above the point where the rod is hinged to the wall...

3. The Attempt at a Solution
I have no clue where to begin...sorry

You'll need to start with a force diagram for this construction. Find the horizontal and vertical components of these forces. Since this is a statics problem, the sum of the horizontal components and the sum of the vertical components are each zero. This will give you essential relations between some of the forces that you'll need in the final solution.

You then need to find the torques caused by all the forces acting on the rod. (You will want to make the hinge your reference point, as this will eliminate some of the unknown forces from the torque equation -- their moment arms will be zero.) The sum of these torques is also zero, since the rod is not rotating. This third equation should give you enough information to solve for all the unknowns.

A uniform beam is 5.0 m long and has a mass of 56 kg. In the figure below, the beam is supported in a horizontal position by a hinge and a cable, with angle ? = 50°. In unit-vector notation, what is the force on the beam from the hinge?

This problem is in much the same vein as the previous one.