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Rzqust [24]
4 years ago
11

The drawing shows three objects rotating about a vertical axis. The mass of each object is given in terms of m0, and its perpend

icular distance from the axis is specified in terms of r0. Rank the three objects according to their moments of inertia, largest to smallest.

Physics
1 answer:
photoshop1234 [79]4 years ago
7 0

Answer:

I₁ > I₃ > I₂

Explanation:

Taking the pic shown, we have

m₁ = 10m₀

m₂ = 2m₀

m₃ = m₀

r₁ = r₀

r₂ = 2r₀

r₃ = 3r₀

We apply the formula

I = mr²

then

I₁ = m₁r₁² = (10m₀)(r₀)² = 10m₀r₀²

I₂ = m₂r₂² = (2m₀)(2r₀)² = 8m₀r₀²

I₃ = m₃r₃² = (m₀)(3r₀)² = 9m₀r₀²

finally we have

I₁ > I₃ > I₂

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A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is  \alpha =2.538 \  rad/s^2

Explanation:

From the question we are told that

  The mass of the wheel is  m =  6.9  kg

   The radius is  r =  0.69 \  m

    The radius of gyration is  k_G = 0.4\  m

    The angle is  \theta = 47^o

    The force which the massless bar is subjected to F = 22.5 \  N

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that  point A is the  center of rotation.

  Generally the moment of  inertia about A is mathematically represented as

     I_a =  I_G + M* r^2

Here I_G is the moment of inertia about G with respect to the radius of gyration  which is mathematically represented as

    I_G =  M *  k_G

=>I_a = k_G*  M + M* r^2

=>I_a =0.4 * 6.9  + 6.9 * 0.69^2

=>I_a =6.045 \  kg \cdot m^2

Generally the torque experienced by the wheel  is mathematically represented as

       \tau =  F *  cos (47)

=>     \tau =  22.5 *  cos (47)

=>     \tau =  15.34 \ kg \cdot m^2 \cdot s^{-2}

Generally this torque is also mathematically represented as

     \tau = I_a * \alpha

=>   15.34  =  6.045 * \alpha

=>   \alpha =2.538 \  rad/s^2

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