How does decreasing the radius change the angular velocity?
A tire is moving with a certain angular velocity. The radius of the tire is decreased. The angular velocity would certainly be affected. Is it increased or decreased? And why? angular velocity= distance moved/ time taken. If the distance is reduced, the time taken would be less too... so angular velocity, in that case would increase. But, if we don't consider the time, distance is reduced so angular velocity is reduced. Im confused. Please answer, thank you :)
Does the velocity increase when the radius is increased in circular motion?
Yes if the centripetal force acting on the body is constant. The formula that governs uniform circular motion is [math]F=\frac {mv^2}{r}[/math]So you can see that to keep F constant, velocity must increase if radius increases.To understand it intuitively look at it like this: when radius increases less force is needed to create the curved path. Since force is still constant, the extra bit of force which is not used to change the direction of the body is used to increase the body's speed. NOTE that this is NOT a scientific explanation, it's just a way of intuitive imagination.
Ok this one is just annoying. I was pretty positive with myself...?
Of course the new force required is four times that initially required that's because the relation comes out to be F= (m*u*u)/(2*d) where u is initial velocity therefore force is proportional to the square of initial velocity hence doubling the velocity will quadruple the force required.
If the earth shrinks to half of its radius without change in mass, the duration of the day will be what?
Without sweating the details, this problem is like the ice skater pulling their hands in towards their body -- conservation of angular momentum says angular velocity increases due to change in rotational inertia (I). I goes like M R^2, so since the new R is 1/2 the original, the new I must be 1/4 the original. As a result, the angular velocity must increase by a factor of 4 to conserve angular momentum. As a result, the day will be 1/4 of 24 hours (6 hours.)
If the Earth shrinks to half its radius, what would be the duration of the day?
It depends on whether or not Earth’s mass shrinks with its radius. Let’s assume that the mass stays the same, so that the Earth just compresses as its radius shrinks. We’ll also make the simplifying assumption that Earth is a sphere of uniform density.Angular momentum is given by:[math]L = I\times\omega[/math]where [math]I[/math] is a quantity called moment of inertia (for some damn fool reason), and [math]\omega[/math] is angular velocity (approximately [math]2\pi[/math] radians per 24 hours for planet Earth).Angular momentum is conserved, so that:[math]L_{i} = L_{f}[/math][math]I_{i}\times\omega_{i} = I_{f}\times\omega_{f}[/math]For a solid sphere, [math]I \propto r^2[/math][math]r^ 2 \times\omega_{i} = (\frac{r}{2})^2\times\omega_{f}[/math][math]4\ \omega_{i} = \omega_{f}[/math]So planet Earth would rotate about four times in 24 hours, reducing our day length to six hours.
If you double the earth in size and mass, will the escape velocity remain the same? Why?
This from Wikipedia article “Escape velocity”:For a spherically symmetric, massive body such as a star or planet, the escape velocity for that body, at a given distance, is calculated by the formula[3] [math]v(e) = sqrt(2GM/r)[/math]where G is the universal gravitational constant (G ≈ 6.67×10−11 m3·kg−1·s−2), M the mass of the body to be escaped, and r the distance from the center of mass of the body to the object.Now (and I add this note) the orbital velocity is 1/sqrt2 times the corresponding escape velocity at the same radius or altitude. The difference of say a hundred miles above the surface of the earth doesn’t make a big difference in orbital velocity.However, in space launches and trajectories, the figures for the orbital elements necessary to describe the vehicle’s motion sufficiently accurately to conduct docking maneuvers or a flight to the moon require seven to sixteen digit accuracy. There are a minimum of six orbital elements required, in either polar or Cartesian coordinates, such as x, y, z and velocity vectors in the x, y and z axes at a given instant. See Wikipedia article “Orbital elements.”The timing accuracy required for space flight is very high; presently a system of antennas called the Deep Space Network (DSN) yields 0.05 mm/second velocity and 3 meters range accuracy for objects being tracked. Timing is coordinated with a Deep Space Atomic Clock (DSAN) which maintains precision of less than 1 nanosecond drift per 10 days.Following is an example of a typical 2-line element used by NASA/NORAD: 1 27651U 03004A 07083.49636287 .00000119 00000-0 30706-4 0 2692 2 27651 039.9951 132.2059 0025931 073.4582 286.9047 14.81909376225249