5)))) A thin, 60.0 disk with a diameter of 9.00 rotates about an axis through it

Question

5)))) A thin, 60.0 disk with a diameter of 9.00 rotates about an axis through its center with 0.260 of kinetic energy. What is the speed of a point on the rim? 4))At time a grinding wheel has an angular velocity of 27.0 . It has a constant angular acceleration of 34.0 until a circuit breaker trips at time = 2.00. From then on, the wheel turns through an angle of 440 as it coasts to a stop at constant angular deceleration. part a Through what total angle did the wheel turn between and the time it stopped? part b At what time does the wheel stop? part c What was the wheel's angular acceleration as it slowed down? 9 ))))During most of its lifetime, a star maintains an equilibrium size in which the inward force of gravity on each atom is balanced by an outward pressure force due to the heat of the nuclear reactions in the core. But after all the hydrogen "fuel" is consumed by nuclear fusion, the pressure force drops and the star undergoes a gravitational collapse until it becomes a neutron star. In a neutron star, the electrons and protons of the atoms are squeezed together by gravity until they fuse into neutrons. Neutron stars spin very rapidly and emit intense pulses of radio and light waves, one pulse per rotation. These "pulsing stars" were discovered in the 1960s and are called pulsars. Part A A star with the mass and size of our sun rotates once every 32.0 days. After undergoing gravitational collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.200 . By treating the neutron star as a solid sphere, deduce its radius. Part B What is the speed of a point on the equator of the neutron star? Your answer will be somewhat too large because a star cannot be accurately modeled as a solid sphere.

Explanation / Answer

Ek = (1/2) I w^2 ==> w = sqrt .26*2/i = .02 I = (1/2)M R^2 I = .5*60*4.5^2 = 607.2 v = wR v = 0.02*4.5 = 0.13m/s

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