1. The sun currently has a mass of 1.99 × 10 30 kg, a radius of 6.96 × 10 8 m, a

Question

1. The sun currently has a mass of 1.99 × 1030 kg, a radius of 6.96 × 108 m, and rotates about once every 30 days. In a little over 5 billion years, our sun will collapse to a white dwarf approximately 16,000 km in diameter. Assume the sun does not lose any mass in this process. Treat the sun as a perfect sphere.

a) How many days will it take to complete one rotation after it has collapsed to a white dwarf?

b) What will the sun’s kinetic energy be after it has collapsed? Was mechanical energy conserved?

2. A uniform steel beam has a mass of M. On it is resting half of an identical beam. What is the vertical force exerted on the longer beam by each support? Express your answers in terms of M and l.

3. A steel cable is designed to support a large industrial light fixture with a mass of 270 kg. You want a safety factor of 10, that is you want the cable to be able to support 10 times as much weight as you will actually have. Steel has a Young’s Modulus of 180 × 109 N/m2 and an ultimate tensile strength 860 × 109 N/m2.

a) What is the minimum cross-sectional area of the steel cable if it is to support a mass of 2700 kg? Assume the cable hangs straight down.

b) If the cable is 7.5 m long, how far does it elongate when supporting the 270 kg light fixture?

4. A brick rests on the bottom of a swimming pool, at a depth of 2 m. The brick has a density of 2400 kg/m3, and dimensions of 25 cm x 10 cm x 8 cm.

What is the buoyant force acting on the brick?

What normal force does the bottom of the swimming pool exert on the brick?

What is the water pressure at the bottom of the swimming pool?  

PLEASE SHOW ALL WORK                                         

Explanation / Answer

1) a) Angular momentum of the sun is conserved about its axis of rotation since no external torques are involved.

I11 = I22

=> (2MR12/5)1 = (2MR22/5)2

=> 2 = 1(R1/R2)2 = (1/30) * [(6.96 * 108) / (8000 * 103)]2 = 252.3 rotations per day

Days taken to complete one rotation, T = 1/2 = 3.96 * 10-3 days

b) Kinetic energy of the sun after collapse,

KE2 = I222/2 = MR2222/5

=> KE2 = (1.99 * 1030) * (8000 * 103)2 * [252.3 * 2 / (24 * 3600)]2 / 5 = 4.3 * 1040 J

KE1/KE2 = (I112/2) / (I222/2) = 1/2 = (1/30) / 252.3 = 1.32 * 10-4 < 1

So, KE1 < KE2

Hence, mechanical energy is not conserved.

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