1) A horizontal disk with a radius of 17 m rotates about a vertical axis through its center.
The disk starts from rest and has a constant
angular acceleration of 2.6 rad/s^2.
At what value of t will the radial and tangential components of the linear acceleration
of a point on the rim of the disk be equal in
magnitude? Answer in units of s.
2) Given: G = 6.67259 x 10^-11 Nm^2=kg^2
After a supernova explosion, a star may undergo a gravitational collapse to an extremely dense state known as a neutron star, in which all the electrons and protons are squeezed together to form neutrons. A neutron star having a mass of 1.5 x 10^30 kg (about equal to that of the Sun) would have a radius of about 9.3 km.
Find the free-fall acceleration at its surface.
Answer in units of m/s^2
a)Find the weight of a 72.9 kg person at its
surface. Answer in units of N.
b)Find the energy required to remove a particle
of mass 1.17 x 10^-27 kg from its surface to
infinity. Answer in units of J.
1.
Let the time be t
So, the angular velocity at the rim at the time (ω) is equal to αt, where α is 2.6 rad/s2
The radial acceleration at this time = ω*ω*R = α2t2*17 = 2.6*2.6*17*t*t = 114.92 t*t.
The tangential acceleration at this time = αR = 2.6*17=44.2 m/s2
So,
t*t = 2.6
t = 1.61 seconds
2)
The acceleration due to gravitation at the surface is given by
GM/R2 = 6.67259 * 10^-11 * 1.5 * 10^30 / (9300 * 9300) = 1.15*10^12 m/s2
a)
Weight = mass * acceleration at surface = 1.15*10^12*72.9 =8.436*10^13 N
b)
The energy required is equal to the change in potential energy, which is equal to the potential energy at the surface, as the potential energy at point of infinity is zero.
Potential = GMm/R = 6.67259 * 10^-11 * 1.5 * 10^30 * 1.17*10^-27 / (9300) = 1.259 *10^-11 Joules
http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation
http://en.wikipedia.org/wiki/Gravitational_potential_energy#Gravitational_potential_energy
http://en.wikipedia.org/wiki/Angular_Acceleration
Report Spam/Abuse