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beks73 [17]
3 years ago
13

How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five

times smaller
Physics
1 answer:
Keith_Richards [23]3 years ago
6 0

Answer:

<em>The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

<em></em>

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = I

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = 5I  (five times larger)

angular speed = ω/5  (five times smaller)

Recall that the kinetic energy of a spinning body is given as

KE = \frac{1}{2}Iw^{2}

therefore,

for the first case, the K.E. is given as

KE = \frac{1}{2}Iw^{2}

and for the second case, the K.E. is given as

KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2}   = \frac{5}{50}Iw^{2}

KE = \frac{1}{10}Iw^{2}

<em>this is one-tenth the kinetic energy before its spinning characteristics were changed.</em>

<em>This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

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gizmo_the_mogwai [7]

Answer:

The velocity just before hitting the ground is v_f = 30 m/s

Explanation:

From the question we are told that

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      v^2 =u^2 + 2as

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Now we are told that was through horizontally with a speed of

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Substituting values

       v_f = \sqrt{10^2 + 28.3^2}

       v_f = 30 m/s

   

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3 years ago
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Answer:

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Explanation:

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