Please HELP! 1.)A horizontal disk with a radius of 18 m rotates about a vertical
ID: 1335853 • Letter: P
Question
Please HELP!
1.)A horizontal disk with a radius of 18 m rotates about a vertical axis through its center. The disk starts from rest and has a constant angular acceleration of 6.5 rad/s2.
At what time will the radial and tangen- tial components of the linear acceleration of a point on the rim of the disk be equal in magnitude?
Answer in units of s.
2.) A m = 52.2 kg object is released from rest at a distance h = 1.0487 R above the Earth’s surface.
The acceleration of gravity is 9.8 m/s2 . For the Earth, RE = 6.38×106 m, M = 5.98 × 1024 kg. The gravitational acceleration at the surface of the earth is g = 9.8 m/s2.
Find the speed of the object when it strikes the Earth’s surface. Neglect any atmospheric friction.
Caution: You must take into account that the gravitational acceleration depends on dis- tance between the object and the center of the earth.
Answer in units of m/s.
Explanation / Answer
1.
radial component = tangential component
w2r = ar
w2 = a
(w1+at)2 = a
a2t2 = a
t2 = 1/a
t= sqrt(1/a) = sqrt(1/6.5) = 0.39s
2.Here we use law of conservation of energy
KEf+PEf = KEi+PEi
1/2mVf2+ (-GMm/Rf) = 1/2mVi2+ (-GMm/Ri)
Vf2 = Vi2 +2GM(1/Rf-1/Ri)
But Ri= 1.0487Rf
Vf2 = Vi2 +2GM(1/Rf-1/1.0487Ri)
Vf2 = Vi2 +2GM/Rf *(1-1/1.0487)
Vi=0m/s
Vf2 = 2GM/Rf *(1-1/1.0487)
Vf = sqrt [2GM/Rf *(1-1/1.0487)] = sqrt[(2*6.67*10-11*5.98*1024)/(6.37*106)* (1-1/1.0487)] = 2411.6 m/s = 2.4km/s