From the diagram in the link you posted you should be able to write an expression for the elongation (or compression) of the spring Δx in terms of the small angular displacement of the wheel from the equilibrium position. The force applied at the point of connection between the spring and the...
Look here
https://www.physicsforums.com/showthread.php?t=110015
to find the 1-D kinematics equation that relates the change in velocity to a known acceleration over a known distance.
A "flat" battery is more complicated than a resistor. The simplest way to model a dying battery is as a constant source of emf with an increasing internal resistance. A voltmeter with huge internal resistance draws essentially no current, so if the voltage is measured with no other load on the...
The angular velocity is not constant. The instantaneous power is changing during the acceleration phase. You cannot use torque times final velocity to find the average power. The average power is the total work done on the wheel divided by the time interval it takes to do the work. The work...
Your initial idea for finding the angular velocity of the system after the collision is correct. Once you have found ω you have two ways to approach the solution. You can either find the center of mass of the whole system and use the distance from the pivot to the CM with ω to find the...
You calculated the velocity of the end of the rod. The only thing moving with that velocity is the tiny bit of mass at that end. Everything else is moving slower.
The linear momentum of a system of particles is the sum of the linear momenta of each bit of mass in the system. The concept of...
All the information you need is given. You need to find the relative change in the wavelength to determine the relative change in velocity, and from that determine the relative change in tension, and ultimately the relative change in the mass.
This does not look right
dV = adt - V_0 = V_1 - V_0 \cr
The definition of acceleration is just a = dV/dt
Since the mass of the rocket is not changing, you can treat the collision//expulsion of a mass dm using conservation of momentum. It does not matter that the gas expelled...
The key point here is that the force of the rope acting on the bike is NOT contributing to the force needed to keep the bike from moving down the plane. There are equal and opposite forces acting on the bike at the ends of the rope between the wheel and the frame. This is no different from...
I assume your part f) is actually asking you to compare the altitude to that attained in part e), since part a) has nothing to do with altitude, but the problem is poorly stated since it does not tell you what to assume about friction. The answer depends on what you assume about the coefficient...