# Tag Info

3

It is easier to explain in terms of torque and angular momentum, but it can also be seen with some difficulty using only internal forces. Imagine that instead of a continuous rod you have instead a bunch of particles attached by cords. If you push the mass at the edge this one will tend to move up first, but due to the cord will also try to rotate around ...

3

The total torque about some axis is defined as: $\vec{\tau}_\text{tot}=\Sigma \left(\vec{r}_i \times{}\vec{f}_i\right)$ If you change to a parallel axis located $\vec{r}$ away from the first one on the plane of the forces, the new torque will be: $$\vec{\tau}_\text{tot}'=\Sigma \left[(\vec{r}_i+\vec{r}) \times{}\vec{f}_i\right] =\vec{\tau}_\text{tot}+\vec{... 2 The question answered by @SatwikPasani is related but not quite a duplicate. The apparent paradox is resolved by realising that using a frame of reference relative to the sphere which is accelerating down the slope is a non-inertial frame of reference. If there is friction and the no slipping condition is satisfied then the frame of reference attached to ... 2 In such a hypothetical situation in which there is no friction between the sphere and plane, there can be no tangential force acting on the sphere, and hence no torque. The only force acting on the sphere would therefore be its weight, and the component of that force acting perpendicularly to the plane would be responsible for its translation down the plane, ... 2 Think of a single particle with mass m at \vec r from a coordinate origin. Now suppose this body is in circular motion with angular velocity \omega around this origin. The momentum of the particle is then$$ \vec p = m{\vec v} = m\frac{d\vec r}{dt}.  The velocity is then $\vec v = \vec\omega\times\vec r$. Now let us consider the angular momentum of ...

2

Your problem cannot be solved without further information. This is because the board and pins in your model are ideal, not real : they are perfectly flat, rigid and inflexible. As an ideal problem it is statically indeterminate : there are infinitely many solutions which are compatible with the given conditions. It is possible to increase some reactions ...

2

In this case, the rod cannot experience only pure translation. Because resultant torque acting on the rod isn’t equal to zero. If acting points of the forces are fixed, the rod will rotate counter clockwise until it is parallel with the forces. So, until that final state of the rod is established (without damping forces, this state will never be ...

2

If you accept that no external work was done, then if there is a change in the state of a system through which the kinetic energy changed, there must be a corresponding change in potential energy. The key to understanding the (rather poorly narrated) video is that the lecturer implies (at T=2:30) that $\Delta E=0$ from which it follows that $\Delta KE= - \... 2 if the body is rolling on a plane, then its degree of freedom is 2: one for rotation about the body's axis and one for translation of its center of gravity in forward and backward direction. if there is no slipping, the translation can be calculated from the rotation and the radius of the block. Thus the degree of freedom is degenerated to 1. 1 Assume that the rod is made by some particles (or molecules). The force$F$is acting on one particle of the rod as you see in the figure above. If we isolate that particle, we will have figure below (free body diagram): Force$f_1$is applied by the particle next to the first particle. As you see in the figure, there is a distance between$F$and$f$... 1 For a pulley that has mass and moment of inertia, there must be a net torque on the pulley for the pulley to demonstrate an angular acceleration. Assuming that mass "M" is greater than mass "m", a net torque necessarily requires that the counterclockwise torque from mass "M", given by the equation Torque1 = T1(R), is larger than the clockwise torque from ... 1 Yes. Tension can vary if external forces are acting between the ends of the string - such as gravity (if the string has mass) and friction where the string makes contact with other objects (such as the pulley). For example, suppose you attach one end A of a uniform massless string to a support and the other end C to a vertically hanging mass M. This ... 1 The tension in a string is uniform Not always. The tension of the string is uniform in some cases: If String is mass-less and its particles don't move with respect to each other (i.e. string is inextensible or if it is extensible, it reach to its final tension). If String is mass-less and there is no friction between string and pulley. For string with ... 1 The torque supplied by$F_m$results in the torque due to$F_v$, so these torques are equal :$\vec {JV} \times \vec F_v = \vec {JM} \times \vec F_m$. Evaluation : (a) either$\vec A \times \vec B = (AB \sin\theta) \hat k$where$A$,$B$are magnitudes and$\theta$is the angle between (b) or$(A_x \hat i+A_y\hat j) \times (B_x\hat i+B_y\hat j) = (A_xB_y ...

1

Any rigid body is composed of particles bonded together. This bonding allows forces applied on one part to transferred to all other parts of the body. Thus when a net force is applied, all the particles will translate together at the same time. We describe this motion is a translation of the center of mass. In addition, the bonded particles try to maintain ...

1

Consider the midpoint of the rod that divides the rod into a left half and a right half. We are applying a force to the right end in the upward direction. Now consider the left half. It has to (overall) go in the direction of the force applied - the rod doesn't break, right? So if we think of just the left half then there must be some net force on it in the ...

1

I think the confusion is that a single push at the corner doesn't "feel" like you're "doing a rotation". Here's an explanation of why it must be, by symmetry. Suppose that pushing it up on the right end, as shown in your diagram, gives the rod angular velocity $\omega$ and linear velocity $v$. (We have not assumed $\omega \neq 0$, we're going to show that.)...

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