# Tag Info

1

Their paper is inconsistent. They filled in $\omega = 264$ with the other quantities in SI units, so ω should be expressed in rad/s (often written $\mathrm{s}^{-1}$). So they assumed ω was already in rad/s. If they say they assumed $\omega = 264~\mathrm{rpm}$, that's not consistent with the values they plugged in. Your value of 69696 is hard to decipher ...

0

Think of a non homogeneous rigid body. It is rotating about a axis, you have to find say Moment of inertiaabout the axis. So in that case you consider a axis for which it's easy to calculate, and which is parallel to the axis you have to find. For examples, for the axes having origin at the centre of mass, then you can find the M.I easily, then use that ...

3

The tangential acceleration $a_t$ and the angular acceleration $\dot{\omega}$ are basically the same thing. They are related by: $$a_t = r\dot{\omega}$$ So we don't include both of them because that would be counting the same thing twice.

0

There has been an apprehension about whether the cyclist can take a turn without steering. In my opinion he can. There are two principles he can use. These are By shifting the line of normal reaction sideways to the line of force acting upon the center of gravity. By trying to rotate the cycle sideways. For simplicity consider the case of a standing man ...

0

you have to calculate the moment of inertia about the centre of mass first, which is, let us call this $I_{CM}$, using the parallel axis theorem you get $I=I_\text{CM}+md^2$ where $d$ is the distance to the centre of mass from where you are calculating the moment of inertia, and $m$ is the whole mass so from the point $x=0$: $I_0=I_\text{CM}+m(4/3)^2$ ...

1

Firstly let's look at the case of a horizontal plane first. In my answer here I derived that the critical friction coefficient is: $$\large{\mu_c=\frac{FI}{mg(I+mR^2)}}$$ Now we have three scenarios: a) No friction at all, $\mu=0$: Assuming no forces or couples act on the object then Newton tells us that the state of motion remains unchanged, or: ...

0

What happens is not a transfer of kinetic energy from the weight to the pulley. Rather, the pulley accelerates slower and then reach a lower speed because part of the work done by gravity is directly transformed into rotational energy of the pulleys. Yes, both translational and rotational motion energy are forms of kinetic energy.

0

Everything in classical mechanics, momentum, angular momentum, torque, velocity etc. is measured about a point. Period. You can be sort of a Newtonian Nazi and complain that it is wrong to talk about torque about an axis and you'll be correct but here it means a completely different thing but in common language, we often make do with such words. So, coming ...

0

Before you read the rest of my answer, you must know that strictly speaking, we always calculate torque about a POINT. But, there is a way to find torque about an axis, provided that axis is the axis of rotation. But, when you find torque about a point, you don't even need any rotational motion. Since you have trouble understanding what torque about a point ...

0

Torque never acts over an axis it acts only at the point of contact whereas moment of inertia acts along an axis

0

It's like any balancing problem. You constantly move your point of support to cause yourself to fall one way or the other. If you don't like the way you are falling, you move your point of support to stop that fall, and then start falling the other way. Your point of support is never stationary. If it is, you fall over. On a bike, you move your point of ...

1

With regard to question B1.3, one can immediately eliminate options b and e because energy is a scalar quantity, something without direction. One can also eliminate option c because the work of art has non-zero kinetic energy from the perspective of the (implied but obvious) observer frame of reference. That leaves options a, d, and f. The second conditions ...

27

They do! There's an entire class of galaxy, called a 'satellite galaxy' which is defined entirely based on them orbiting a larger galaxy (which would be called a 'central galaxy'). Our own milky-way is known to have many orbiting satellite galaxies, or at least 'dwarf-galaxies'. If dwarf-galaxies aren't enough, the milky-way itself is gravitationally ...

90

There are plenty of satellite galaxies orbiting larger galaxies. The question is how long are you willing to wait for an orbit? The Milky Way has a mass $M$ of something like $6\times10^{11}$ solar masses, or $10^{42}\ \mathrm{kg}$. The small Magellanic Cloud is at a distance $R$ of $2\times10^5$ light years, or $2\times10^{21}\ \mathrm{m}$. A test mass ...

0

The moment of inertia is definitely affected by where the axis of rotation is located. To find the torque required to rotate an object where the axis of rotation is not through the center of mass, you definitely need to use the parallel axis theorem. If you intend to rotate a real-world object in such a fashion, expect a lot of "wobble" in the object if ...

2

Let you apply force $\bf F$ at point $P$ the coordinate of which is $\bf r$ measured from a specific point $O$ - the point about which you want to rotate. Let $\bf r$ and $\bf F$ be in the same plane. Now, if you were to rotate $P$ about $O$, it would rotate around some axis perpendicular to the plane in which the force and the point lies; if ...

0

HINT: Density is an intrinsic property of a body.It is independent of how much of the material is present and is independent of the form of the material, e.g., one large piece or a collection of small particles. Here the mass is based on area of the disc and not the volume. Let the density of the disc be equal to $d$. Then, Mass of total disc initially ...

1

Because the work done by friction is converted into rotational kinetic energy of the cylinder, since friction provides the torque to roll down the cylinder.

1

When you are changing gears you are trading speed for torque (or vice versa). The overall power transmitted maintains the same so $P=\omega_I T_I = \omega_O T_O$. The way this works is by the chain forcing the same tangential velocity on the two sprockets (input and output sprocket) from which their angular velocity is found $\omega_I = \frac{v}{r_I}$ and ...

0

I believe that firstly we need to get one thing straight and that is, on infinitesimal level, where is the acceleration of a body moving on some circular path pointing? The answer is, it is always pointing away from the center of curvature of your path, road, or whatever. So then, there is always a force in action, force which is trying to move you ...

1

If you are sliding across the surface, then "static friction" is not applicable. Consider first your motion on a merry go round without sliding. At any instant, your tangential velocity is the same as the tangential velocity of the surface under your feet. Since the two velocities are the same, no instantaneous frictional force is required to keep you moving ...

0

You have two bodies $m_1$ and $m_2$ placed with distances $d_1$ and $d_2$ from some arbitrary point A to their center of masses, then combined mass moment of inertia at that point is $$I_A = (I_1 + m_1 d_1^2) + (I_2+m_2 d_2^2)$$ The combined system has mass $m_1+m_2$ and if the required mass moment of inertia about the combined center of mass C is $I_C$ ...

0

There is a easy semi-geometrical way of finding the center of rotation due to a force. Find the moment arm $c$ of the force through A. $$c = r \cos \theta$$ Find the radius of gyration about the center of mass C $$\rho = \sqrt{ \frac{I_C}{m} }$$ Measure the distance $\ell$ away from the center of mass and mark point R $$\ell = \frac{\rho^2}{c}$$ Point ...

0

I think what you are asking about is answered by the fundamental theorem in the mechanics of rigid bodies, which states that the motion of any rigid body can be decomposed into the motion of its center of mass (not necessarily rectilinear) and a rotation about its center of mass (COM). The two statements you emphasize are direct corollaries. Please see ...

0

Remember subscripts! To avoid confusion. For object $S_2$ you find the expression $T=mg$. But remember to write it as $T_2=m_2g$ since there are more $T$'s and $m$'s in this system. Now find $T_1$ in the same way as you found $T_2$. You have already explained the forces acting on $S_1$. (As @Vishwaas points out in a comment, the ramp's angle is needed here ...

1

Yes, rotational velocity and acceleration is shared by all points on a rigid body. We only state that a body rotated about a point because the linear velocity is zero at that point. See related answer here: http://physics.stackexchange.com/a/215165/392

0

About the only relationship worth considering is whether the planet orbits close enough to its parent star so that tidal forces lock the rotation period to the orbital period. Even this is fraught with problems because we currently don't know the exact "tidal friction" coefficients for exoplanets. This will depend greatly on the structure of these planets ...

4

Rotational speeds of planets cannot be calculated/predicted because planet formation seems to be highly chaotic. The spin of planets (both rocky and gas) is determined by many factors, including: the angular momentum of the material which was accreted on the planet, gravitational interactions with other planets, the history of collisions as the ...

1

Check out this chart as very rough baseline. Solar energy, even with 88 days of Mercury level sunshine, Wouldn't reach nearly as far into Mercury as you suggest. A few KM, perhaps 10 or 20, but not 4,000. I could back that up with a thermal energy calculation of Mercury's mantle and compare it to annual solar energy it gets hit by, but I'm quite sure ...

1

Be careful with your integration elements, as $dm=\rho dA$, where $\rho$ is the area density of the plate and $dA=dxdy$ is the area integration element. In addition you should integrate from the rotation point to the end of your plate. So for example; $$I_{xx}\\=\int_A(y^2+z^2)\rho dA\\=\rho\int_{x=0}^{0.3\text{ m}}dx\int_{y=0}^{0.6\text{ ... 0 Wait...wait... what? No, no, no, no... No. If a body is spinning, supposing non frictional surfaces and all of that, the energy would not decrease (if there is no external force). The energy is always constant. In that kind of problems there is only rotational energy (there is no other energy):  E_{rot}={1\over 2}I\,\omega^2, ... 1 Yes, the rotational kinetic energy decreases. The extra energy is converted to thermal energy in the wheel and environment. If you imagine letting the weight go, it will slide across the surface of the wheel as it moves towards the edge. This sliding is motion against friction, so energy is lost there. Then the weight might bang into whatever holds it at ... 0 The energy is dissipated when the mass stops at the perimeter. Work has to be done to stop the radial motion of the mass. 0 The fact that it rotates on one end without sliding probably has something to do with the shape of the ruler and how it is in contact with the table. Presumably, the middle part of the ruler is in good contact, and the frictional force on the ruler at that point provides a pivot point allowing it to rotate. If you push it hard enough, you are likely ... 1 If the collision is inelastic, try this:$$mvh = I_m\omega_m + I_M\omega_M$$I_M, the moment of inertia of the solid sphere, is taken as that about its tangent at the point of contact with the ground.$$I_M = \frac {7MR^2}{5}$$and, since the particle of mass m revolves about the centre of the sphere,$$I_m = \frac {mR^2} {2} Here, the angular ...

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