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3

I know this is an old thread, but I had to figure this out for a problem on my physics homework. What helped me to understand this is to think about 2 objects on a spinning disk, one being close to the center of the disk and one being close to the outside of the disk. Angular (rotation) speed deals strictly with the angle. How long does each object take to ...


0

First Thought (probably not the fastest) Let us assume you have a vector space in $R^{3}$ with a quaternion defined as: $$ \mathbf{q} = q_{F}^{*} \ q_{B} \\ = a + b \hat{\mathbf{x}} + c \hat{\mathbf{y}} + d \hat{\mathbf{z}} $$ where $(a, b, c, d)$ are the Euler parameters and $(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}})$ defines the reference ...


0

In order for an object to travel on a curved path of any type (circular or otherwise), it must experience a component of acceleration perpendicular to the path. The magnitude of that sideways acceleration must be, instantaneously, $$a_{\perp}=\frac{v^2}{r},$$ where $r$ is the instantaneous radius of curvature of a circular path, and $v$ is the instantaneous ...


2

There is a very detailed analysis of the physics of spinning balls, torque and drag, in http://www.cim.mcgill.ca/~mpersson/docs/AeroSphere_Persson.pdf More simply, according to this earlier answer, in the case of a sphere spinning in place the torque experienced is given by $$ \vec\tau = -8\pi R^3\eta\vec\Omega $$ but that result seems to be valid only ...


4

The coefficient of static friction and the normal force together allow you to calculate the maximum force that friction can apply, not the actual force. A block sitting on a shelf with no external horizontal forces will have a friction force of zero. If you apply a force from the side that is less than the maximum friction force, then the actual friction ...


0

Direction of frictional force is perpendicular to the motion. Coefficient of friction cannot exceed certain value. If the centripetal force normal to motion exceeds the permissible value, car will skid. The equation you are using will measure speed for maximum frictional force. You are assuming that it does not change, which is not the case. Coefficient of ...


0

The thing is that you cannot measure friction between the tires of the car and the road and then calculate its speed.There are two kind of friction. The first is the static friction and the other is kinetic friction. the maximum static friction is measured by $f_s \le f_{smax}\left( = \mu_s mg\right) $ In order for the car not to slip and make the turn, ...


-2

Clearly by logic if we will go lower than the velocity obtained we can make an easy turn.....So we need to find the max velocity .. If u will increase Ur velocity by this max velocity then the two tires will lift from the ground and then Ur car will get damage.


2

The answer by Bryson S. is solid, thorough and vey good, as is Jerry Schirmer's hint. This is merely another way of looking at the problem. We can consider, as Jerry Schirmer points out, two components of acceleration; a tangential and a normal component. Before we begin, note that velocity always points tangential to the path a particle travels. This is ...


8

Here is one simple way. A point is moving around a circle. It has a blue position vector and a red velocity vector, like this: The position vector stays the same length and rotates around and around in a circle. Because the position vector is changing, it has a derivative. That derivative is the velocity. Because we're always going the same speed, the ...


3

Imagine you are going in a circle of radius $r$ starting at three o'clock and heading towards two o'clock. If it takes you time $T$ then your speed is $v=2\pi r/T.$ Now if you look at your velocity it starts out going upwards then ends up going left, then down, then up again. Its like your velocity vector is on a circle of "radius" $v$ but starting at 12 ...


0

For the radius of the circle, $$ R = \frac{x^2 + y^2}{2 y}, $$ where $x$ is the horizontal distance you want to traverse and $y$ is the vertical distance you want to traverse. Proof: Let's assume the Roomba is at the origin, and facing in the positive $x$-direction. The circle it will traverse is of radius $R$ and is centered at $(0,R)$. This means that ...



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