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Minchanka [31]
3 years ago
11

Determine experimentally which rotational axis yields the maximum rotational inertia (i.e., moment of inertia) and which yields

the minimum rotational inertia for the broom stick. Draw a picture of the broom stick with its axis of rotation for (i) the minimum rotational inertia and (ii) the maximum rotational inertia.
Physics
1 answer:
Law Incorporation [45]3 years ago
3 0

Answer:

  the maximum is I₁ axis of rotation at the end

     the minimum moment is I₂ axis of rotation at the center of mass

Explanation:

For this exercise we use the definition moment of inertia

          I = ∫ r² dm

for bodies of high symmetry it is tabulated; In this case we can approximate a broomstick to a thin rod, the moment of inertia with respect to a perpendicular axis when varying are

at one end

           I₁ = ⅓ mL²

in in center

           I₂ = \frac{1}{12} m L²

There is another possible axis of rotation around the axis of the broom, in this case we have a solid cylinder

           I₃ = \frac{1}{2} m r²

remember that the diameter of the broom is much smaller than its length, therefore this moment of inertia is very small

when examining the different moments of inertia:

     the maximum is I₁ axis of rotation at the end

     the minimum moment is I₂ axis of rotation at the center of mass

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An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a
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Answer:

16.25^{\circ}

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Horizontal range is given by

R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}

The angle at which the arrow is to be released is 16.25^{\circ}.

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3 years ago
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3 years ago
Ricardo, of mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg canoe. When the canoe is at r
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Answer:

m=57.65 kg

Explanation:

Given Data

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Canoe mass m₂=30 kg

Canoe Length L= 3 m

Canoe moves x=40 cm

When Canoe was at rest the net total torque is zero.

Let the center of mass is at x distance from the canoe center and it will be towards the Ricardo cause. So the toque around the center of mass is given as

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We have to find m₂.To find the value of m₂ first we need figure out the value of.As they changed their positions the center of mass moved to other side by distance 2x.

so

2x=40

x=40/2

x=20 cm

Substitute in the above equation we get

m_{x}=\frac{m_{1}(L/2-x)-m_{2}x }{L/2+x}\\m_{x}=\frac{80(\frac{3}{2}-0.2 )-30*0.2}{3/2+0.2}\\m_{x}=57.65 kg

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