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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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7 0
3 years ago
Read 2 more answers
Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

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