5 Qustions Center of mass(CM)

[匿名]

2009-11-02 2:34 pm

1.)A fishing rod consists of three uniform 80-cm Lenghts with mass 10g, 20g &amp;
30g.locate the CM with respect to the end of the 30g section

2.)where is the CM of the earth- moon system relative to the center of the earth

3.)Find thee kinteic energy of the CM motion and the kineic energy relative to theCM in the following cases (a.)5-kg block moves at 12m/s toward a 1kg block at rest; (b.) a 1 kg block moves at 12m/s toward a 5kg block at rest.

4.)a 2kg particle has velocity (2i-6.2j)m/s and a 3.6kg particle has a velocity
(0.75i+1.2j)m/s. Determine: (a.)the total momentum, (b.)the velocity of the CM.

5.)a 75kg man is at one end of the uniform 25kg platform of length 4m.intially at rest. He can walk at 2m/s relative to the platform, which can roll freely.
(a.)where is the CMof the system initially to the man?(b.)what is the speed of the platform when the man walks to the other end? (c.)what is the distance moved by the platform when he reaches to the other end?

please show at the work

回答 (1)

Prof. Physics

2009-11-03 4:15 am

✔ 最佳答案

1. We can treat the CM of the three rods are located at their centre respectively.

That is, the CM of the 30g, 20g and 10g are 40 cm, 120 cm and 200 cm from the end of the 30g rod respectively.

Let x be the required distance

x = [(30)(40) + (20)(120) + (10)(200)] / (30 + 20 + 10)

= 93.33 cm

2. The mass of the Earth is 81 times that of the Moon.

Assume x be the distance between the CM relative to the centre of the Earth.

The distance between the Moon and the Earth is 3.8 X 10^8 m

x = M(3.8 X 10^8) / (81M + M)

= 4.63 X 10^6 m

which is inside the Earth.

3.a. Speed of the CM = 5(12) / (5 + 1) = 10 m/s towards the 1 kg block

Kinetic energy relative to the CM = 1/2 (5)(12 - 10)^2 + 1/2 (1)(-10)^2 = 60 J

b. Speed of the CM = (1)(12) / (5 + 1) = 2 m/s towards the 5 kg block

Kinetic energy relative to the CM = 1/2 (1)(12 - 2)^2 + 1/2 (5)(-2)^2 = 60 J

4.a. Total momentum

= mu + Mv

= (2)(2i - 6.2j) + (3.6)(0.75i + 1.2j)

= 6.7i - 8.08j

b. Velocity of the CM

= (mu + Mv)/ (M + m)

= (6.7i - 8.08j) / (3.6 + 2)

= 1.20i - 1.44j

5.a. Initially, distance between CM and the man

= (25)(2) / (25 + 75)

= 0.5 m

b. By conservation of momentum

Initial momentum = Final momentum

0 = (75)(2) + 25v

Velocity of the platform, v = -6 m/s

So, the speed is 6 m/s

c. As there is no external force acting on the system, the CM must not move.

Let d be the distance moved by the platform

So, d - 4 is the distance moved by the man relative to the ground.

0 = [(75)(d - 4) + 25d] / (25 + 75)

d = 3 m

參考: Physics king

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