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nata0808 [166]
2 years ago
14

a disk initially at rest rotates with constant angular acceleration through an angel of 50 rad in 5 sec find the angular speed o

f the disk just after 3 sec. please step by step
Physics
1 answer:
lisov135 [29]2 years ago
3 0

Recall the formula,

∆<em>θ</em> = <em>ω</em>₀ <em>t</em> + 1/2 <em>α</em> <em>t</em> ²

where ∆<em>θ</em> = angular displacement, <em>ω</em>₀ = initial angular speed (which is zero because the disk starts at rest), <em>α</em> = angular acceleration, and <em>t</em> = time. Solve for the acceleration with the given information:

50 rad = 1/2 <em>α</em> (5 s)²

<em>α</em> = (100 rad) / (25 s²)

<em>α</em> = 4 rad/s²

Now find the angular speed <em>ω </em>after 3 s using the formula,

<em>ω</em> = <em>ω</em>₀ + <em>α</em> <em>t</em>

<em>ω</em> = (4 rad/s²) (3 s)

<em>ω</em> = 12 rad/s

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Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim
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75.36 mph

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(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

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z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)

The rate of change is,

2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)

2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots

Now finding z when t=0, from (1) we have

z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}

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The officer's radar gun indicates 25 mph pointed at the other car then, \frac{d z}{d t}=25 when t=0, from

From (2) we get

2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)

2(0.7071)(25)=V+2 V^{2}(0)-40

35.36=V-40

V=35.36+40=75.36

Hence the speed of the car is 75.36 mph

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