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4

Assuming rolling without slipping, this can easily be solved by means of Energy Conservation. Let's assume the incline has a height $h$. During the travel down the incline, potential energy $U$ is then converted to kinetic energy $K$: $K=U$ $K=mgh$ The kinetic energy $K$ is a combination of translational and rotational energy: $\frac12mv^2+\frac12 I\...


4

The flaw is that you've failed to do an experiment which will tell you whether the frame is inertial or not. If you do such an experiment -- for instance take a test mass, initially at rest with respect to the frame, release it, and see if it remains at rest -- you will immediately discover that the frame is not inertial.


3

The equations of motion for the position determine the accelerations: they are second-order differential equations in time: $$\vec F = m\vec a = m\ddot{\vec x }$$ So the acceleration, the second derivative of the location in time, has to be determined from the state of the physical system in some way. Typically, it is determined using the $F=ma$ formula ...


3

This is an excellent question which is much more subtle than it first appears. On a first (or even a second) reading, you seem to answer your own question and then ignore your own answer! What I mean is, you clearly understand that while in an accelerating vehicle you experience a 'jerk' which does not happen with uniform motion, and you also acknowledge ...


3

The object at the end of the rope has a negative velocity (in the world frame of reference) when the elevator (and it) comes to a halt. When the elevator stops, you have to go from a negative velocity to zero velocity. That requires a positive acceleration.


3

Indeed your first suggestion is wrong :$ \Delta x = v_o t + gt $ Instead it should be $ \Delta x = v_o t + gt^{2} $(You can recheck it) Where you are wrong is here: According to your question v is the final velocity since $(v=v_{0}+gt)$ So $\Delta x\neq vt$ but instead it should be $\Delta x =v_{average}\times t$ In uniform acceleration $v_{average}$ ...


2

When we talk about kinematics of a point particle, mainly the velocity and acceleration vectors, it would be reasonable to use differential geometry to understand what is going on. The motion of a point particle is described fully if it's given its position vector $\:\mathbf{r}(t)\:$ as function of time $\:t\:$. Then the velocity vector $\:\mathbf{v}(t)\:$ ...


2

For a car going around a corner of a constant radius moving with a constant speed the magnitude of the centripetal acceleration will be constant but the direction of the acceleration will change. So we would have to say that the centripetal acceleration is not constant in the same way that the velocity is not constant.


2

The centripetal acceleration will always be directed towards the center of the circular arc that the car's instantaneous path is a part of. Thus, the direction of the acceleration will be along the line joining car's position and the point about which it is in instantaneous circular motion. We can easily visualize that during a turn, this line is changing ...


2

The mass of the wheel is located farther from the axis of rotation, thus the moment of inertia is greater for the wheel. So it rotates slower with the same rotational energy and therefore the ball rolls faster down the slope than the wheel.


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A body in an accelerated reference frame (say, a train with acceleration $\mathbf{ A}$) will appear subjected to an inertial force, i.e., it's necessary to add $\mathbf{F}_i=-m\cdot\mathbf{A}$ to the 'real' forces acting on the body for Newton's second law to hold in this reference frame. You can obtain this result from a change of reference frame, which, ...


2

The solution is actually strange for me. It assumes that the distance traveled by each car (in the CM frame) during the acceleration (crumpling) is also equal to the crumple length of each car, but is not always the case. Following the assumption of the solution: In the CM frame, the heavier car moves slower than the lighter car. So that at the moment that ...


2

The flaw is your assumption that In this [accelerating] frame we don't see any force so the first law of dynamics is respected. In the accelerating reference frame you do see evidence of a force, even though you don't see the effect you are expecting (acceleration of the object). Like an observer standing on the surface of the Earth, acted on by the ...


2

You're probably used to thinking about the objects at the bottom of ropes pulling downward on ropes because of gravity. But here, it's better to think of the top of the rope pulling up. This is because here's the sequence of events when the bottom of the elevator hits the ground: 1) The ground exerts a large upward force on the elevator floor in order to ...


1

Roughly, this means that if the spacecraft travels 1000 inches on the first day, it will slow down enough to travel only 700 inches on the second day, and then slow down enough more to travel only 400 inches on the third day, and then slow down enough more to travel only 100 inches on the fourth day. In other words, it takes the spacecraft one day to slow ...


1

They ask you for the angular acceleration $\alpha$, not the linear acceleration. I could not understand what you did, but you know that torque $\tau=I \alpha$, to find $\alpha$ only divide the torque you have by the moment of inertia $I=mL^2/12$ which corresponds to the moment of inertia of a rod around its center of mass.


1

This is a simple question of conceptual understanding. The only force acting on the ball after it is released from the hand is that of the ball's weight due to gravity. Since it is the only force, the consequent acceleration is also the only acceleration. Gravity is being taken to be $10 m/s^2$ in a downward direction. For the answer to be correct, our ...


1

Let's take the first equation of motion which is : \begin{equation} v=u+at \end{equation} Integrate this equation to get: \begin{equation} \int\frac{dx}{dt}dt=\int{u}dt+\int{at} dt \end{equation} this gives: \begin{equation} x=ut+\frac{1}{2}at^2+x_0 \end{equation} The integration constant can be done away by putting the proper limits on $x$.(Assuming the ...


1

The FLRW energy equation for the motion of test masses in the universe is $$ \left(\frac{\dot a}{a}\right)^2 = \frac{8\pi G\rho}{3}. $$ the scale factor for space is $a$ and its time derivative is $\dot a$. I derived this from Newtonian dynamics. The density of mass $\rho$ for the case of a quantum vacuum energy level is constant. I now replace this with ...


1

According to the definition of the work ($\delta W=F\mathrm dx$), it is better that we say friction doesn't do work. Because, at each point of contact area, friction force is fixed and doesn't move. Friction converts some portion of energy used to move the box, to heat.



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