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6

You only need one force, applied off the object's center of mass. Such a force produces rotational motion, as you know, but it must also produce linear motion. You deliver some amount of linear momentum to the object over the time you are applying the force (because that is what forces do). If there is no other force on the object that can take that momentum ...


4

The statement, "Objects don't rotate when they fall" is false. I know this as an experimental fact b/c I worked on SRTM (https://www2.jpl.nasa.gov/srtm/). Consider the object in the figure: We can approximate this as an asymmetric dumbbell of length $L=60\,$m and two masses, $M=2\,$Gg (the space shuttle), and $m=20\,$Mg (the outboard radar ...


4

Any net force that is not directly in line with the COM of an object, that doesn't have a fixed axis, will cause both spatial translation and rotation of the object. For the object to have translation only, the force must be directly in line with the COM. To have rotation only on the object, equal and opposite forces would have to be applied on equally ...


3

In most cases, when a traction force applied to the rear wheels exceeds the friction, they start sliding sideways (in regard to their intended track). The rear part of the car tries to follow a straight trajectory outside the intended turning curve. Front wheels, on the other hand, have enough friction to follow their track (at least some more time, if the ...


3

What you are referring to are not called gyroscopes (which usually refer to sensors in this context), but reaction wheels or momentum wheels (not the same device). Reaction wheels spin the craft in the opposite direction that the wheel is spinning, while momentum wheels are a spinning wheel where the axis of rotation is mounted on a gimbal and when the ...


3

The thing is that the relation $a_t=\alpha r$ gives the tangential component of the acceleration $a$, i.e. $a_t=a\sin{\theta}$. You can see this by differentiating $\vec{v}=\vec{\omega}\times \vec{r}$. You'd get $\vec{a}=\vec{\alpha}\times \vec{r} + \vec{\omega} \times \vec{v}$. The second term is directed along $\vec{r}$ and is called radial acceleration. ...


2

The relationship $\tau = I \alpha$ is the correct one. It relates torque about an axis to the angular acceleration about the same axis, and the mass moment of inertia also about this axis. It is true that if $\tau$ is a result of an offset force $F$ at a radius distance $r$, with angle $\theta$ between the force direction and the radial direction, then $$ \...


2

You are correct in finding this exception. Whether this occurs will (obviously) depend on the actual values of $M$, $m$, $\mu$ and $u$, as well as the sizes of the blocks. By solving the initial differential equation, you can obtain a condition (inequality) for which this will never happen (where the relative velocity reaches zero before the slippage reaches ...


2

You can calculate the torque with respect to any given point in any given coordinate system. In particular, you can calculate the torque relative to the center of mass with all vectors represented in the inertial coordinate system. Only because you choose some point to calculate the torque with respect to, doesn't mean you have to change the coordinate ...


2

It has to do with the sum of the moments at the reactions. There are two requirements for the bridge to be in equilibrium. The sum of the forces and the sum of the moments have to be zero. $$\sum \vec F=0$$ $$\sum \vec M=0$$ Assuming the only load on the bridge (ignoring the weight of the bridge) is the person walking across, then the sum of the reactions at ...


2

No, what you describe is not a possibility. To explain why let me make a comparison with the technology of using gyro stabilization in luxury yachts. The gyrostabilizer is oriented with respect to the hull in such a way that it is set up to suppress rolling. In order to suppress rolling the gyrostablizer needs leverage. A boat, being elongated, has strong ...


2

You can do it trivially. Imagine a point mass $m$ at $\vec r_0$. At rests. Now apply a force $\vec F$ for time $\Delta t$. You can shrink $\Delta t\rightarrow 0$ and keep the impulse fixed: $$\Delta \vec{p} = \vec F \times \Delta t$$ The particle then has momentum $\vec p$ and velocity $\vec v$. It has also acquired angular velocity about the origin: $$ \vec{...


2

In real situations a pulley will have mass and frictions. So string tension will have to become higher on one side of the pulley than the other side to provide a net torque high enough to overcome frictions and start rotating the mass of the pulley.


2

It makes sense, because the larger $l$ is, the larger the moment due to $mg$ about the point that static friction (which always balances $mg$) acts. This moment is trying to turn the rod counter-clockwise. As you've discovered, the location of the normal force must shift to balance this moment for static equilibrium. That is, $N$ provides a clockwise moment. ...


1

Let $\vec M$ be the torque in 3D-space, $\vec r$ the position vector of some mass point $m$ and $\vec F$ the force applied to $m$, then it holds $$ \vec M = \vec r \times \vec F~. $$ Now assume the third components $r_z$ and $F_z$ are both 0, then there follows $M_x = M_y = 0$, too (just write those components of the vector product explicitly), and you are ...


1

But watching car videos or such, people also say "10 ft-lbs OF TORQUE".....isn't that redundant? It is necessary to specify torque since ft-lbs is also a unit for work which is not the same thing as torque. Hope this helps.


1

It is because pound-feet correspond to torque, but not only torque. Energy has the same physical dimension of $\mathrm{kg \cdot m^2 \cdot s^{-2}}$ or $\mathrm{lbs \cdot ft^2\cdot s^{-2}}$, yet they are not directly comparable. In case of torque this is the result of a cross product of two vectors $\vec{r} \times \vec{F}$ (length of the lever arm multiplied ...


1

It's convention and a judgement about whether the meaning will be clear if we state a quantity without also stating its meaning. We also might say the engine has "2 liters displacement" (rather than just "2 liters"), or "200 braking horsepower" (rather than just "200 hp"). On the other hand if we say a milk jug has &...


1

Yeah, it's because of the other useful quantities which have those units.


1

The horizontal frictionless bearing is an axle going into the paper, through the center of the circle. We are looking at it from the side, but the pulley wheel will have some thickness. It's like a bike wheel and somehow designed (with rolling bearings, or lots of oil) so there is no friction when it turns. The details aren't too important for the question, ...


1

A gravitational field affects every bit of matter equally - for any object on earth, the acceleration due to gravity is 9.8 $m/s^2$. When an object is at rest, there is clearly no relative movement between any parts of the object, which would be observed as a bending, stretching, or rotation of the object. When you drop that object, every bit of it is ...


1

The main reason I have ran into is so that the wider front forks on a motorcycle do not hit the fuel tank which is generally located close to the front forks. Also most motorcycles have a "triple tree" type front end which has to have more frame clearance than the typical bicycle steering stem needs.


1

It is better to consider only the forces that are perpendicular to the symmetry plane. Shown above at the forces acting on the top arm of the tool. To crush the nut (and to answer your first question) the torque about the pivot due nut must be less that the applied torque due to the hand $$ x N < (x+y) F $$ But to estimate the crushing force, solve the ...


1

What does it mean by "the ends of this cut will pull apart"? How does "pulling apart" look like? Why would "the ends of this cut will pull apart"? Actually, the statement was "the ends of this cut will pull apart if the force acts outward from the boundary". For a beam or wire that would mean the beam or wire is in ...


1

The dipole translates if there is a difference in field strength (in magnitude or direction) over the length of the dipole, and it rotates if it is not aligned with the electric field. It follows that the dipole will purely rotate about its centre in a uniform field if it is not aligned with the field vector, and it will purely translate in a non-uniform ...


1

You can think it this way. There are 2 forces with non zero torques: The weight force $W$ acting on the centre of gravity Reaction force $R$ from the cube acting horizontally Since the board is in rest, those torques should compensate each other. Torque is the force (red) times arm of the forces (green). We have an equation: $$ W\cdot \frac x2 = R\cdot d. $...


1

Imagine stretching or contracting the bridge. This changes the total length $L$ by some factor $L \to \lambda L$. Intuitively the bridge will still be in equilibrium even after this rescaling. This implies that the equilibrium condition cannot depend on the total length of the bridge, but only on the relative position of where the various forces are applied. ...


1

Simplest solution: The lower mass m is struck by a horizontal impulse giving it a large momentum (mu). It leaves so quickly that the upper block does not gain a significant angular velocity. The upper block experiences a horizontal impulse from friction: μMgt = Mv where (t) is the contact time. It drops to the lower surface where it receives a similar (...


1

The $\sin(\theta)$ is included in the definition of the torque, which uses only the component of the force which is doing work in the tangential direction.


1

To elaborate on R.W. Bird's excellent and correct answer: if the masses are not suspended precisely in a horizontal line with the suspension point of the bar, then the horizontal position is a stable equilibrium configuration rather than a neutral equilibrium configuration. To see this, let's first draw a "thick" rod in horizontal equilibrium: ...


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