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dmitriy555 [2]
1 year ago
13

A device consisting of four heavy balls connected by low-mass rods is free to rotate about an axle. It is initially not spinning

. A small bullet traveling very fast buries itself in one of the balls. In the diagram, m = 0.005 kg, v = 450 m/s, M1 = 1.2 kg, M2 = 0.5 kg, R1 = 0.8 m, and R2 = 0.3 m. The axle of the device is at the origin < 0, 0, 0 >, and the bullet strikes at location <0.274, 0.752, 0> m. Just after the impact, what is the angular speed of the device? Note that this is an inelastic collision; the systems temperature increases. angular speed (absolute value of the angular velocity) = _________ radians/s
Physics
2 answers:
zubka84 [21]1 year ago
4 0

The angular speed of the device is 1.03 rad/s.

<h3>What is the conservation of angular momentum?</h3>

A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.

Using the conservation of angular momentum

L_{i}=L_{f}

Here,  = the system's angular momentum before the collision

L_{i} = 0 + mv

= (0.005)(450)(0.752)

= 1.692 kgm²/s

The moment of inertia of the system is given by

I = 2(M₁R₁² + M₂R₂²)+ mR₁²

= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²

= 1.6292 kgm²

Here,  = Iω

So,

1.692 = 1.6292(ω)

ω = 1.03 rad/s

To know more about the conservation of angular momentum, visit:

brainly.com/question/1597483

#SPJ1

nordsb [41]1 year ago
3 0

The angular speed of the device is 1.03 rad/s.

<h3>What is the conservation of angular momentum?</h3>

A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.

Using the conservation of angular momentum
L_{i} = L{f}

Here, L_{i} = the system's angular momentum before the collision

L_{i} = 0 + mvd_{y}

= (0.005)(450)(0.752)

= 1.692 kgm²/s

The moment of inertia of the system is given by

I = 2(M₁R₁² + M₂R₂²)+ mR₁²

= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²

= 1.6292 kgm²

Here, L_{f} = Iω

So,

1.692 = 1.6292(ω)

ω = 1.03 rad/s

To know more about the conservation of angular momentum, visit:

brainly.com/question/1597483

#SPJ1

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Answer:

v = 33.66 m/s

Explanation:

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now time taken by sound to come back at the position of shooter is given as

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7 0
3 years ago
A weight suspended from a spring is seen to bob up and down over a total distance of 20 centimeters twice each second.
Tom [10]

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6 0
3 years ago
A ball is pushed into a spring-loaded launcher with a force of 20 N, which compresses the spring 0.08 m.
damaskus [11]

Answer:

0.8J

Explanation:

Given parameters:

Force  = 20N

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Unknown:

Spring constant  = ?

Elastic potential energy  = ?

Solution:

To solve this problem, we use the expression below:

           F = k e

F is the force

k is the spring constant

e is the compression

             20  = k x 0.08

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Elastic potential energy;

       EPE  = \frac{1}{2} k e²    =  \frac{1}{2}  x 250 x 0.08²

 Elastic potential energy = 0.8J

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