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3 years ago

7

The uniform 12 ft pole is hinged to the truck bed and released from the vertical position as the truck starts from rest with an

acceleration of 3 ft/sec^2.
If the acceleration remains constant during the motion of the pole, calculate the angular velocity ω of the pole as it reaches the horizontal position.

1 answer:

Anna35 [415]3 years ago

7 0

Answer:

w = 0.886 rad / s

Explanation:

Angular and linear variables are related

a = α r

where a is the linear acceleration, α the angular acceleration and r the radius of gyration

α = a / r

the angular velocity we can find it

w² = w₀² + 2 α θ

the initial angular velocity is zero, the angles to be horizontal is

θ = π/ 2 rad

we substitute

w = √ 2 a / r θ

we calculate

w = √ (2 3/12 π/2)

w = 0.886 rad / s

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(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh

makvit [3.9K]

Answers:

(a) $2509.98 m/s$

(b) 397042.215 m

(c) 1917.76 m/s

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is $700 km$ and whose gravitational acceleration at the surface is $a_{g}=4.5 m/s^{2}$

Knowing this, let's begin:

a) In this part we need to find the escape speed $V_{e}$ on the asteroid:

$V_{e}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the universal gravitational constant

$M$ is the mass of the asteroid

$R=700 km=700(10)^{3} m$ is the radius of the asteroid

On the other hand we know the gravitational acceleration is $a_{g}=4.5 m/s^{2}$ , which is given by:

$a_{g}=\frac{GM}{R^{2}}$ (2)

Isolating $GM$ :

$GM=a_{g}R^{2}$ (3)

Substituting (3) in (1):

$V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R}$ (4)

$V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)}$ (5)

$V_{e}=2509.98 m/s$ (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

$E_{o}=E_{f}$ (7)

Being:

$E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R}$ (8)

$E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h}$ (9)

Where:

$E_{o}$ is the initial mechanical energy

$E_{f}$ is the final mechanical energy

$K_{o}$ is the initial kinetic energy

$K_{f}=0$ is the final kinetic energy

$U_{o}$ is the initial gravitational potential energy

$U_{f}$ is the final gravitational potential energy

$m$ is the mass of the object

$V=1510 m/s$ is the radial speed of the object

$h$ is the distance above the surface of the object

Then:

$\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h}$ (10)

Isolating $h$ :

$h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R$ (11)

$h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m$ (11)

$h=397042.215 m$ (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance $h=981.8 km=981.8(10)^{3} m$ . This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

$\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2}$ (13)

Isolating $V$ :

$V=\sqrt{2a_{g} R(1-\frac{R}{R+h})}$ (14)

$V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})}$ (15)

Finally:

$V=1917.76 m/s$

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Explanation:

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Finally, between 4 s and 6 s, the position goes from -4 m to 6 m. The displacement is:

Δx = 6 m − (-4 m)

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The net displacement is the change in position from 0 s to 6 s:

Δx = 6 m − 0 m

Δx = 6 m

In the second part of problem 1, we have a velocity vs time graph.

Car 1 starts with 0 velocity and ends with a velocity of 6 m/s, so it is accelerating and constantly moving to the right.

Car 2 starts with a velocity of -6 m/s and ends with a velocity of 6 m/s. It is also accelerating, but first it is moving to the left, comes to a stop at t = 3 s, then moves to the right.

Car 3 starts with a velocity of 2 m/s and ends with a velocity of 2 m/s. So it is moving constantly to the right, but never speeds up or slows down.

We want to know when two of the cars meet. Unfortunately, this isn't as easy as looking for where the lines cross on the graph. We need to calculate their displacements. We can do this by finding the area under the graph (assuming all the cars start from the same point).

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So we know Car 1 and Car 3 meet, we just have to find where and when. For Car 1, the area under the curve is a triangle. So its displacement is:

Δx = ½ t v(t)

where t is the time and v(t) is the velocity of Car 1 at that time. Since the line has a slope of 1 and y intercept of 0, we know v(t) = t. So:

Δx = ½ t²

Now look at Car 3. The area under the curve is a rectangle. So its displacement is:

Δx = 2t

When the two cars have the same displacement:

½ t² = 2t

t² = 4t

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t (t − 4) = 0

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t = 0 refers to the time when both cars are at the starting point, so t = 4 is the answer we're looking for. Where are the cars at this time? Simply plug in t = 4 into either of the equations we found:

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