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allochka39001 [22]

3 years ago

13

A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2/3. Suppose a 2.50 m ro

d with a mass of 3.00 kg is hinged at one end and is held in a horizontal position. The rod is released as the free end is allowed to fall. What is the angular acceleration as it is released?

1 answer:

tatiyna3 years ago

4 0

Answer:

6 rad/s²

Explanation:

Sum the torques about the hinge.

∑τ = Iα

mg(L/2) = mL²/3 α

g/2 = L/3 α

α = 3g/(2L)

α = 3 (10 m/s²) / (2 × 2.50 m)

α = 6 rad/s²

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jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

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Where:

$y=0m$ is the final height of the ball

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$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

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Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

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Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

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$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

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If a book has a mass of 3 kg, what is the book's weight in N?

sesenic [268]

Answer:

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juin [17]

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For second case

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at highest Point velocity is $u\cos \theta _2$

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as there is no acceleration in x direction therefore horizontal velocity is same

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Gnoma [55]

Answer:

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