We’re rewarding the question askers & reputations are being recalculated! Read more.
9

It is a common mistake among introductory physics students to assume that the coefficient of static friction $\mu_s$ determines the strength of the static friction force $F_s$ as expressed through the equation $F_s=\mu_sN$, where $N$ is the normal force interaction between the two surfaces in question. This equation is actually false except for a very ...


6

Resistance against the rotation greatly depends on the point of suspension. Hanging the spindle on a hook will always provide lesser resistance than hanging it straight on the rod. When you hang the spindle on a hook, the point of suspension is in line with the axis of rotation. When it is not so, there will be a component of string's tension that would ...


5

As both the ball and the car roll down the incline their potential energy is converted into kinetic energy. There are two forms of kinetic energy, linear and rotational kinetic energy. The linear kinetic energy gain determines how soon they reach the bottom of the incline. Now for the ball, all of its mass is involved in the rotation, while only the wheels ...


4

The angle subtended $\theta$ is varying with the x. So whenever x changes theta will change. Since x is a function of time, it depends on time. But theta depends on x, and it is clear from that theta depends on time. In $x=sin (\theta)$ ,$ \theta$ is the variable and while we taking the derivative with respect to time, $\theta$ should be considered. If $\...


4

Looks very much like the formula at this link, except that wedge product is used instead of vector product and unit vectors instead of directional derivatives EDIT(11/01/2019): So let me derive the formula. It should be noted that it was FUBAR by an editor, who omitted plus signs in the following original formula of the OP: $$\mathbf{\omega}=\frac{1}{2}\...


3

The equation $$\boldsymbol\tau=\mathbf r\times\mathbf F$$ Is the definition of the torque $\boldsymbol\tau$ of a force $\mathbf F$ that is applied at position $\mathbf r$ relative to some "reference point" (where $\mathbf r=0$). The equation $$\boldsymbol\tau=I\boldsymbol\alpha$$ Is an analog of Newton's second law ($\mathbf F=m\mathbf a$). Therefore, in ...


2

you can use this equation for a friction torque in a spherical joint. $$\tau_\mu=-\text{signum}( \dot{\theta})\,\mu\,|F_N|\,R$$ where $\tau_\mu$ is the friction torque $\mu$ friction coefficient $F_N$ normal joint force $R$ Joint radius the "signum" function is necessary, because the friction torque must operate always opposite to the angular ...


2

If you make a free body diagram for the system you will realise that without friction it is impossible for the system not to move. You don't even need to consider the torque. If you apply Newton's 2nd law you can see it. $$\sum_i \overrightarrow{F_i} = \overrightarrow{F_{B, 1}}+\overrightarrow{F_{B, 2}}+\overrightarrow{W}+\overrightarrow{N}=m\...


2

Your intuition is exactly right. In particular, the torque "from gravity" is actually coming from the friction at the point of contact. Because we're worried about rotation about the center of mass of the object (as it rolls), gravity itself cannot provide a torque (because the lever arm is zero). So gravity only acts on the center of mass and generates only ...


2

The two bodies exchange (scalar) momentum $p$ along the contact normal $\boldsymbol{n}$, and through the contact point $\boldsymbol{r}$ (relative to the center of mass). Here is a schematic of the situation Linear momentum is inherently balanced because whatever momentum vector $p\, \boldsymbol{n}$ is added to (2) it is subtracted from (1). $$ m_1 \...


2

Well, their moment of inertia are the same if the weight distribution of whorl in both the cases is the same. So, any torque given, should in theory, produce same net change in angular momentum. So, if you are doing this with a machine, you would need to spend same amount of electricity(work done) to spin a particular amount of yarn. I think your question ...


2

I was wondering how the stationary flywheel scenario worked. As Adrian said if the flywheel were not to spin, it would fall over (i.e. you would not observe what is called precessional motion). Why? Because there's no angular momentum due to the spin of the flywheel and thus no torque. Why is it that the gyroscopic falls along a "circular path?" This is ...


2

Your treatment is odd. Your band (the ring) will make sense if the axis is vertical (labeled as "X" on your figure), because it then will have great symmetry about the vertical axis. For the horizontal axis, your band does not have good symmetry so you will have difficulty integrating it. Specifically, your $dI$ does not equal to $dm x^2$ because different ...


1

Since theta is also a function of time, you need to apply the chain rule. Angle is variable due to the horizontal motion of arm OP. Regardless, the very fact that they are asking for the first and second derivatives of angle implies that is non-constant in nature, else they would be zero.


1

Torque is best defined as the work per unit angle (as in Joules per radian) that can be done by a force which is acting in a manner which might cause a rotation. As we learned in statics, it can be calculated relative to any point in your system by using the expression r x F.


1

Conceptually, torque is the turning effect of a force. It is dependent upon two factors- one is the size and direction of the force, and the other is the distance between its point of application and the point around which it is causing the rotation. In the example you give, the rotation of the wheel is actually a series of instantaneous rotations about ...


1

It's hard to say whether or not what you talk about is "due to inertia". Inertia is essentially just mass, but I think the term is usually either misused or just unclear when it comes up, so I try to stay away from it. We we can say is that if a ball is floating in space and rotating (as viewed from a non-accelerating reference frame), then it will continue ...


1

A rigid body rotating about its center of mass has angular momentum. You need torque to change this angular momentum just as you need force to change linear momentum. These statements say nothing about mass or inertia on their own $$ \begin{aligned} \boldsymbol{F} & = \tfrac{{\rm d}}{{\rm d}t} \boldsymbol{p} \\ \boldsymbol{\tau} & = \tfrac{{\rm d}}{{...


1

Convention is to use the right hand rule. Your right thumb points in the direction of the torque (moment) vector produced by the rotation associated with the curl of your fingers. Thumb pointing out of the page is positive torque due to a counter clockwise rotation. But as already noted it’s really arbitrary. As long as you are consistent when setting ...


1

Are you asking how to calculate the piston side forces? Like any other problem in mechanics, start with a sketch of the problem with all relevant dimensions. Add the driving degrees of freedom that completely describe the state of the system. For example the crank angle. Split each part into its own free body diagram and describe the applied forces as well ...


1

You can waste energy and power and other things, but you can't really waste force. We don't have a fixed supply of force that runs low. Energy transfers when you apply a force over a distance. If you apply a force and there's no movement, then no energy is required. In your example of a piston converting rotational to oscillatory motion, there are ...


1

The line element in polar coordinates is $\vec{ds}=dr\hat{r}+rd\theta\hat{\theta}$. Divide by dt, you get $\frac{\vec{ds}}{dt}=\frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}$ We are given that $\frac{dr}{dt}=12m/s$ and $\frac{d\theta}{dt}=2$rad/s. We multiply by 8 m to get $r\frac{d\theta}{dt}=16$m/s. The instantaneous velocity is the vector sum of ...


1

Imagine we have a ball rotating around a pole due to a light string (sort of like a tetherball setup). We can calculate the angular momentum of the ball about the pole. If we imagine drag is minimal, then the ball spins quite freely. In this case, since the only force on the ball is from the string, and that force goes through the axis we are ...


1

These are the fundamental definitions you must know and the rest follow from these. Let's say you are in reference frame $S$ with origin $O$. Assume that there's a particle of mass $m$ located at position $\vec{r}$ (called the position vector) with respect to $O$ and its velocity is $\vec{v}$. Then, $$\vec{L}_{\text{of the point particle with respect to a ...


1

In the first case, since the string is assumed to be massless, you can apply Newton's law to it to see that forces on both ends of the string must be equal.


1

The magnitude of the angular momentum of the rod about its center of mass after the impulse will be $Pd$. The linear momentum will be $P$, so the magnitude of the contribution of the center-of-mass motion to the angular momentum about point $O$ will also be $Pd$, but in the opposite direction. Thus the total angular momentum about $O$ is zero both before and ...


1

Will this contraction emit photons? Generally, electrically neutral objects do not radiate when they are subjected to acceleration unless the acceleration highly upsets the uniform distribution of the negative (electrons) and positive (protons) charges inside matter. if a line is fixed on the diameter on the rotating disk, each point on the diameter ...


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