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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?
Physics
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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Answer:

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A biologist looking through a microscope sees a bacterium at r⃗ 1=2.2i^+3.7j^−1.2k^μm(1μm=10−6m). After 6.2 s , it's at r⃗ 2=4.6
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The Average velocity for the bacterium is 0.75 unit/sec.

<u>Explanation:</u>

The given values are in the vector form

Where,

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dT = time interval

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|d S|=\sqrt{d S^{2}}

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As dT=7.2 sec

Now,

Solving the equation, we get;

v(a v g)=\frac{5.390732789}{7.2}

\begin{aligned}&v(a v g)=\frac{5.39}{7.2}\\&v(a v g)=0.748611111\\&v(a v g)=0.75 \text { units / sec }\end{aligned}

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Karolina [17]

Answer:

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