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Minchanka [31]
2 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]2 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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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
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Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

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Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

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     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

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Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

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So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

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Answer:

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