The height h doesn't depends of inclination; it is measured from the ground.
From the drawing you made, it is clear that not all gravitational potential energy goes into compressing the spring, if you take ground as a point of 0 gravitational potential energy. Try to find the difference in...
A spring is comprensed or stretched along only one axis. That said, it seems that the confusion here is what must we use as x. If 12 cm (0.12 m) is the compressed lenght, then x = 0.12 m. But if 0.12 m is the final lenght of the spring, x would be instead 18 cm.
One question: is the spring...
Right, the rope makes an angle of 100° with the body. With this in mind, you can draw a component vector that begins from the body and goes up to the tip of the tension vector, in such way that the component is perpendicular to the body.
You have then a 90°-80°-10° triangle, from which you can...
Yes, friction forces are applied at different distances from the axis. That said, the torque is greater for cilinder A, and it comes first.
One question: is the lower part of cilinder B contributing to normal force? If it isn't, the problem is now solved.
But, if the inner surfaces that are in contact to the two sides (bases) of the cilinder are not frictionless, there would be a reduction of the torque... And it would also reduce kinetic energy...
Anyway, ignoring that, all seems to indicate that they come at the same time.
But, if the inner surfaces that are in contact to the two sides (bases) of the cilinder are not frictionless, there would be a reduction of the torque... And they would also reduce kinetic energy...
Anyway, all seems to indicate that they come at the same time.
If both cilinders have the same normal force (which is evenly distributed in the bits), then the friction force is the same, and they come to the floor at the same time. Cilinder B could be lower than A, but I think we can ignore this.
Gneill, I think that efficiency here refers to energy; in other words, only 82% initially available energy is useful, and the other 18% is converted to other forms of energy.
I would calculate kinetic energy (it is not the same as the potential energy, since the efficiency is less than 100%). Then, I would find speed, and later calculate the speed after the inelastic collision. Then, I would use conservation of energy.
Mmm... do the cilindrical bits have mass? If they don't, then they would not affect the moment of inertia... If you have some diagram or picture you can upload, if would greatly help in understanding the problem :)
You must apply parallel-axis theorem to find the moment of inertia (since the axis doesn't pass through the center of mass). Take in account the displacement of the center of mass and the rotational and translational kinetic energies when applying conservation of energy. Remember this equation...
Another way to solve this problem is to use just two directions for forces. It may sound crazy, but I have seen this procedure in two or more textbooks.
The positive direction would the direction of motion (the direction of falling of the 12-kg block). This way, if m1 is the 12-kg block, and...
To find the components of the weight, draw an angle formed by the vector of weight and a line perpendicular to and under the surface. This angle is equal to the angle of inclination of the plane.
Make the x axis parallel to the surface, so the y axis would be perpendicular to the surface and...
If angular momentum doesn't change, then the product of the initial angular speed and moment of inertia is equal to the product of the final angular speed and final moment of inertia.
Mmmm... I have just tried to solve this problem using the equation of kinematics I gave and the work-kinetic energy theorem and both gave me that the average force is 32000 N. Do you know what is the correct answer (from the textbook, if you took the problem from one)?
About using momentum...
Try using this equation: 2ax = (vf)^2 - (vi)^2.
Find the acceleration.
Use Newton's second law to calculate force.
This is the only possible solution I can see.
Note: use above equation only if a is constant.
I like your approach using work-kinetic energy theorem.
One point: the force...
Ok, the only possible way (at least for me) is performing a triple integral. This integral, however is easy to solve, but a bit tedious.
Make r^2 = x^2 + y^2. Integrate dm = p dV over the region ocupied by the cube (the limits of integration are 0 to L for all three axis). Remember that dV = dx...
The reduction of mass makes the moment of inertia smaller. If angular momentum is conserved, how do you think would angular speed change? It is easy to tell it taking in account what I said above.