New answers tagged rotational-kinematics
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I figured it out, it's quite simple.
I am citing formulas from this reference page:
The formula for the magnitude of the velocity is $\| v \| = r \omega$ where $r$ is the radius defined as the line segment from the origin of $s$ frame to the "fixed point" about which the screw motion is applied (the thick black dot in the picture). Radius = 2. ...
-1
Deriving the relation between angular velocity and linear velocity-
The greater the rotation angle in a given amount of time, the greater the angular velocity. Angular velocity ω is analogous to linear velocity v. We can write the relationship between linear velocity and angular velocity in two different ways: v=rω or ω=v/r.
Linear velocity is simply how ...
-1
We can define the angle as the area sweep by a vector in the rotating plane with its strating point at the rotating position. As the vector rotates a angle $\Delta \theta$, the area swept by the vector is:
$$
\Delta A = r^2 \Delta \theta
$$
Therefore, we may define the angle as the area swept divided by $r^2$. The area is a vector quantity with ...
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the components of the unit vector $~\vec{e}_x~$ in $~x'~,y'$ are
$$\vec{e}_{x'}=\begin{bmatrix}
\cos(\theta)\\
-\sin(\theta)\\
0\\
\end{bmatrix}$$
the components of the unit vector $~\vec{e}_y~$ in $~x'~,y'$ are
$$\vec{e}_{y'}=\begin{bmatrix}
\sin(\theta)\\
\cos(\theta)\\
0\\
\end{bmatrix}$$
and
$$\vec{e}_{z'}=\begin{bmatrix}
0\\
0\\
...
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I think the answers already present give a good idea of what rotation matrices do. Because you expressed specifically confusion about this business of which matrix acts on the basis and which on the components, here is a quick intro to what transfromations do to bases and the components of vectors.
Vectors, Bases and Components
Choose any basis of a vector ...
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Say you have a vector $\mathbf{r}=\begin{pmatrix}x & y \end{pmatrix}^T=x\mathbf{i}+y\mathbf{j}$ in the cartesian coordinate system. If you want to rotate that vector by some angle $\theta$ then the rotated vector $\mathbf{r}^{\prime}$ would be described by
$$\mathbf{r}^{\prime} = A \mathbf{r} = \begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta &...
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First $X=(x,y,z)$ is a vector expressed in a basis $\mathcal{B}=(e_1,e_2,e_3)$ which is probably the standard euclidian basis $\mathcal{B}=((0,0,1) ,(0,1,0),(0,0,1))$ here. $x,y,z$ are the components of the vector $X$ in the basis $\mathcal{B}$.
Actually you can have a rotation matrix around Z axis : $A(t)=\begin{matrix}\cos(t) & sin(t) & 0 \\
-\sin(...
2
Once the forces and their positions are known, its possible to calculate the torque with respect to the COM. The forces can be affected by the acceleration, but supposing that they are measured correctly (by a load cell for example) that effect is already being taken in consideration.
For example an yoyo toy falls at an acceleration $a$ while I keep the end ...
2
For a system of particles, with the origin of the coordinate system taken as the center of mass (CM), $d\vec J/dt = \vec \tau$ where $\vec J$ is the angular momentum with respect to the CM and $\vec \tau$ is the torque from the forces in the inertial frame with respect to the CM. This is true even if the CM is accelerating. The CM of mass is special in this ...
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Individual forces do not change, but you need to add fictitious forces that may or may not contribute to the torque. In the case of an object undergoing linear acceleration, the inertial term can contribute to the torque if your origin of coordinates is not at the center of mass.
1
$$\vec{y'}=\left[ \begin {array}{c} \sin \left( \theta \right) \sin \left( \phi
\right) \\\cos \left( \phi \right)
\\ \cos \left( \theta \right) \sin \left( \phi
\right) \end {array} \right]
$$
$\Rightarrow$
$$\tan(\Phi)=\frac{\sin(\Phi)}{\cos(\Phi)}=\frac{\vec{y'}_x}{\vec{y'}_y}=\frac{\sin(\theta)\,\sin(\phi)}{\cos(\phi)}$$
thus:
$$\sin(\Phi)=\sin(\...
2
This problem is not nearly as simple as it is presented. I'm pretty sure the solution you've linked to is either wrong or hiding some major assumptions. The key seems to be "But the net acc. of A in the vertical direction should be zero," but no justification for that claim is given, and I don't think it's even right.
The question is the ...
0
In the second case, immediately after the second string is cut, the first sting is still vertical and there is no horizontal force acting on the object. The acceleration of the center of mass is vertical and would be the same as the point mid-way between, A, and, B; rather than tangential.
-1
a=rα as i understand gives the acceleration of a single point with angular acceleration α and the distance from axis as r.
As we know that I is of a point from the axis is defined as mr^2 where r is the perpendicular distance from the axis of rotation. Also we know that F=ma
Using this and the equation τ=F∗r=Iα we can easily prove a=rα
We have to keep in ...
2
This is not always true but in case of pure rolling
consider a figure
In case of pure rolling suppose the rim of ring moves a distance x corresponding the ring covers an angle $ \theta$
$$X=R\theta$$
On differentiating
$$dx/dt =R \omega$$
Hence $R\omega$
Further differentiating
$$dv/dt =R \alpha$$
Hence $a=R\alpha$
3
The key difference between Instantaneous Axis Of Rotation(IOAR) and Radius Of Curvature(ROC) is
IOAR: The point about which all points of a rigid body have perpendicular velocities at a
particular instant.
ROC: $R= v^2/a_\mathrm{normal}$ where the velocity is perpendicular to the center of curvature which is at a distance $R$ from the particle whose ROC we ...
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If your sphere is rolling in a straight line on a hard level surface, then there is no torque which would cause a precession. Then s = Rωt.
0
If the location of where you want to measure speed is $\pmatrix{x & y}$ from the instant center of rotation, then
$$ \pmatrix{v_x \\ v_y} = \pmatrix{ -y\,\omega \\ x \, \omega} $$ where $\omega$ is the angular speed.
In the case of a rolling ball, the center of rotation is at the contact point if there is no slipping, and thus the center of the ball is ...
0
Rotational quantities, θ, ω, and α, are usually represented by vectors along the axis of rotation. For a rolling object, they are perpendicular to the translational velocity and parallel to the surface on which rolling occurs. The arc length, s = r θ, does not change if the surface is tilted.
2
Earth has more kinetic energy than the cloud from which it was formed .Where did this energy came from?
Earth has less gravitational potential energy than the cloud from which it was formed. The gain in kinetic energy came from the loss of gravitational potential energy. The process of gravitational collapse also increased the thermal energy of the Earth, ...
0
I'm not sure how you know how much kinetic energy the cloud had, but if you consider the earth alone it is not an isolated system. So energy conservation doesn't apply and as the earth was formed, the cloud was subject to gravitational forces from nearby bodies which could change the amount of energy that was in the cloud to start with. That change could ...
3
It should not create confusion because a magnitude has some units associated to it, but the inverse is not always true, so it can be dangerous to think it that way.
In other words, the units can give you clues of what it is, but it is not the only thing you should look at.
For example, energy is measured in "joules", $1\text J=1\text N\cdot 1\text ...
3
I think it should also be pointed out that
angular velocity is what is known as a vector which means it has both magnitude and direction. Your equation for instantaneous angular velocity, should really be expressed with a a unit vector so that
$$\vec \omega = \frac{ \Delta \theta}{ \Delta t} \hat u$$
where the unit vector $\hat u$ points in the direction of ...
1
The circumference of a circle radius $r$ is $2\pi r$.
So if the circle rotates at $f$ revolution per second, a point on the cirumference moves a distance $2\pi r f$ in one second, so its velocity is $2\pi r f$.
If we measure the rotation speed as $\omega$ radians per second, a point on the circumference moves a distance $r\omega$ in one second.
Since other ...
3
Your book uses the following relation
$$\omega = \frac{d \theta}{dt}$$
One can write ds element as $ds = Rd\theta$ where this becomes
$$\omega = \frac{1}{R}\frac{ds}{dt}$$
So it's in $s^{-1}$ units. It's because radian is a dimensionless unit in SI thus rad/s is equivalent to $1/s$. Frequency $f$ or $\nu$ are in units of $1/s$ and period $T$ is in second. In ...
1
The second method is incorrect. If you calculate how fast the constantly changing contact point is moving, it is only moving horizontally, and the horizontal acceleration can be ignored. If you follow the instantaneous motion of the current contact point, you should get only the centripetal acceleration (it is rotating so the acceleration is orthogonal to ...
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Consider the free body diagram:
Just one force $F$ acts on the body.
To work out whether the object will translate or not, apply $\text{N2L}$, in both $x$ and $y$ directions:
$$F_x=-F\cos\theta=ma_x$$
$$F_y=F\sin\theta=ma_y$$
So there's acceleration ($a$) in both senses.
There's also torque $\tau$ about the CoG, if $\delta$ is the perpendicular distance ...
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It depends: if you give it an impulse aligned with the center of mass, then it will not rotate
However, if the impulse is not aligned (which is what always happens, because impulses are never fine enough), then it will also rotate
It will rotate about the center of mass, of course, because we always like to describe rotations around that point.
However, you ...
-1
This can be judged by observing the motion in future
In due moment of time we see the distance between centre and left rail decreases but in order to ensure that rolling occurs the cone will have to bent as seen in last diagram which indicated it has turned left
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I will give you 2 hints.
Assume that the body never turns .
Then at some distance x calculate linear velocity both cones about the centre.
Hints: 1. in pure rolling linear velocit is given by rw.
W is same for both bodies as it is a rigid body.
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