# Tag Info

Suppose the ramp wasn't there, then the trajectory of the object would the same as if it fell off a cliff: To get the equation of motion you simply note that the horizontal and vertical coordinates are given by (neglecting air resistance): $$x = ut$$ $$y = \tfrac{1}{2} g t^2$$ So you can get the trajectory by substituting for $t$ to get: y = ... 4 Surpringingly the top speed is not necessarily anything to do with friction, though friction will of course have some effect. A motor acts as a generator, i.e. if you turn a motor it will generate a potential difference just like a generator, and this potential difference (usually called the back EMF) is proportional to the motor speed. So if you connect a ... 3 If you assume that Your body is a uniform, thin rigid rod. One end of the rod is pivoted (aka your feet) during the fall. Then one simply recalls that the angular velocity \omega of rotation of your body is related to the tangential velocity v of a point a distance r from the pivot by \begin{align} v = \omega r \end{align} Now, if you have height ... 2 If the bumpers where vertical then the contact point would be at the center and since gravity is more than the bounce force it means there isn't going to be enough friction in change the rotation of the ball when the direction of the ball changes. If the contact point is further up, then the contact force is towards the center of the ball, and hence is ... 2 Good work and a good idea. d = L/2 would correspond to the moment of inertia for a point mass M at distance L/2. Momentum p = M \omega/2 L/2. What would you get if the mass of the rod was concentrated at the two end points, each 1/2 M ? One point zero, the other M/2 \omega L. In other words d = 1. So the mass distribution along the rod plays a role. ... 2 There are a couple of problems with this. One is that\tau_{net}=I\alpha$$is an equation that relates net torque to angular acceleration, but T shown in the figure is a force (tension), not a torque. So use the definition of torque to convert T into torque due to the tension (\tau_{T}). Also, the relationship between the linear acceleration a_{cm} ... 2 Let's suppose I have some system and I know M, the system's total mass, \vec{r}_{cm}, the system's center of mass position and \vec{L}_{cm}, the systems angular velocity in the frame where the center of mass is the origin. How do I find \vec{L}', the angular momentum with respect to some other origin, say \vec{r}_{cm} + \Delta \vec{r}, which is ... 1 I believe that the direction of the friction would actually be different depending where you hit below or above the center of mass. If you hit the ball below the centre the ball still rolls forward but at first the spin is in the opposite direction but changes direction due to the friction and overall forward inertia of the ball. The friction would have to ... 1 John Rennie's answer is great for DC motors but I thought I'd mention why AC induction motors reach a top speed. For AC induction motors, top speed is limited by the speed of the rotating magnetic field set up by the stator (aka, the synchronous speed of the motor). The rotor in an AC induction motor can only rotate as fast as rotating magnetic field and ... 1 There are two parts to angular momentum that both contribute at the same time. In vector form (where × is the cross product)$$ \vec{H}_A = I_{cm} \vec{\omega} + \vec{r}_A \times m \vec{v}_{cm} $$For a horizontal rod rotating about end point A you have$$ \begin{aligned} \vec{\omega} & = (0,0,\Omega) & \vec{v}_{cm} &= \vec{\omega} \times ...