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Katyanochek1 [597]
3 years ago
9

According to the Guinness Book of World Records (1999) the highest rotary speed ever attained was 2010 m/s (4500 mph) The rotati

ng rod was 15.3 cm long. Assume that the speed quoted is that of the end of the rod. a. What is the centripetal acceleration of the end of the rod? (2.64 x 107 m/s2) b. If you were to attach a 1.0 g object to the end of the rod, what force would be needed to hold it on the rod?
Physics
1 answer:
Zinaida [17]3 years ago
7 0

Answer:

a. 2.645 * 10^7 m/s^2

b. 2.645 * 10^4 N

Explanation:

Parameters given:

Velocity of rod = 2010m/s

Length of rod = 15.3cm = 0.153m

Mass of object placed at the end of the rod = 1g = 0.001kg

a. Centripetal acceleration is given as:

a = (v*v)/r

Where v = velocity

r = radius of curvature.

The radius of curvature in this case is equal to the length of the rod, since the rod makes the circular path of the motion.

Hence, centripetal acceleration at the end of the rod:

a = (2010*2010)/(0.153)

a = 26432156.86 m/s^2 = 2.64 * 10^7 m/s

b. The force needed to hold the object at the end of the rod is equal to the centripetal force at the end of the rod. Centripetal force is given as:

F = ma = (m*v*v)/r

Where a = centripetal acceleration

F = 0.001 * 2.64 * 10^7

F = 2.64 * 10^4N

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8. A crate of bananas weighing 3000 N is shipped from South America to New York, where it
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X<br> x<br> 2.5 N<br> 1.7 N<br> 2.5 N
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LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
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Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
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