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julsineya [31]
3 years ago
12

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are

0.847 m and 1.27 kg. What constant-magnitude force acting at the other end of the rod perpendicularly both to the rod and to the axis will accelerate the rod from rest to an angular speed of 6.37 rad/s in 9.87 s?
Physics
1 answer:
igor_vitrenko [27]3 years ago
4 0

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

T = I\alpha = 0.303*0.6454 = 0.196 Nm

So the force acting on the other end to generate this torque mush be:

F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N

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The weight of a boy having a mass of 50 kg is __N.<br> ???
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A 5.20 kg chunk of ice is sliding at 13.5 m/s on the floor of an ice-covered valley when it collides with and sticks to another
Stella [2.4K]

Answer:

The chunk went as high as

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Mass of ice 2 M2= 5.20kg

velocity of ice 1 before impact U1= 13.5 m/s

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3 years ago
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Answer:

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