Hot answers tagged

2

Your left forefinger is exerting a force on one end of the band. Your right forefinger is exerting a force on the other end. Each end of the band is exerting a force on the finger with which it is in contact. You can see this, because the ends of your forefingers have turned white and have temporary grooves in them! [Try not to suppose that the band is '...


2

Consider a mass $m$ in uniform circular motion with radius $r$ and angular frequency $\omega$. It experiences a force of magnitude $F = mr \omega^2$ towards the center of the circle at all times. Its position as a function of time will be something like $(x, y) = (r \cos \omega t, r \sin \omega t)$ (use trigonometry and $\theta = \omega t$ to see this.) At ...


1

From the equation of motion for driven, damped harmonic oscillator results that the amplitude is givn by $$ A=\frac{F_m }{\sqrt{m^2(\omega^2-\omega_d^2)^2 +b^2\omega_d ^2 }} $$ where Fm is the aplitude of the driving force, $\omega_d$ is the driving frequency, $\omega$ is the oscillator's frequency and b is the damping factor. From this we can see that the ...


1

While the spring is going from $\Delta x$ to zero (the normal length at rest), the mass is being accelerated, and is in contact with the spring. Just after reaching the unstreched position, the spring starts to decelerate, and the mass detaches from it. We can use the definition of a conservative force as: $F = -\frac{dE}{dx}$, and calculate the average ...


1

why we use circular motion to explain spring force? It is because the shadow of a body in uniform circular motion experiences a force exactly similar to that experienced by mass on a spring. In fig, the mass $m$ is rotating at a constant angular velocity $\omega$ in a circle of radius $A$. Even though the particle is experiencing a centre seeking force, the ...


1

This is an answer to the heavily edited question. Firstly calculate the velocity of the mass $B$ just prior to impact on $A$, from conservation: $$2mg\times 3x=\frac12 2m v_0^2$$ $$v_0=\sqrt{6xg}$$ The collision is inelastic but without external forces, so conservation of momentum applies. An 'instant' after the collision $A+B$ travel at $v_1$: $$2m\sqrt{...


1

Assume the velocity at impact of $M$ is $v_0$ (which can be calculated from the drop height but that's trivial). The collision is inelastic but without external forces, so conservation of momentum applies. An 'instant' after the collision $M+m$ travel at $v_1$: $$Mv_0=Mv_1+mv_1=(M+m)v_1$$ $$v_1=\frac{M}{M+m}v_0$$ Now calculate the kinetic energy $K$ of $M+m$:...


Only top voted, non community-wiki answers of a minimum length are eligible