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pishuonlain [190]

3 years ago

14

An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that a person inside is stuck

to the wall and does not slide down when the floor drops away. The acceleration of gravity is 9.8 m/s 2 . Given g = 9.8 m/s 2 , the coefficient µ = 0.521 of static friction between a person and the wall, and the radius of the cylinder R = 4.4 m. For simplicity, neglect the person’s depth and assume he or she is just a physical point on the wall. The person’s speed is v = 2πR T where T is the rotation period of the cylinder (the time to complete a full circle). Find the maximum rotation period T of the cylinder which would prevent a 40 kg person from falling down. Answer in units of s.

1 answer:

ElenaW [278]3 years ago

7 0

Let N be the normal force that forces the person against the wall.

Then u N = m g is the frictional force supporting the person's weight

and N = m g / u

also, N = m v^2 / R is the normal force providing the centripetal acceleration

So, m g / u = m v^2 / R

v^2 = g R / u

since v = 2 pi R T

4 pi^2 R^2 T^2 = g R / u and T^2 = g / (4 u pi^2 R)

T = 1/ (2 pi) (g /(u R))^1/2 = .159 * (9.8 m/s^2 / (.521 * 4.4 m)) ^1/2

T = .68 / s

Do you see any thing wrong here?

T should have units of seconds not 1 / seconds

v should be 2 * pi * R / T where T is the time for 1 revolution

So you need to make that correction in the above formula for v.

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To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

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One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at

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Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

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Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is $=P\Delta V$

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$V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L$

$\Delta W=1\times (12.315-2.463)=9.852 J$

$\Delta q=\Delta W=9.852 J$

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In the motion of the medium particles in a longitudinal wave, the medium vibrates parallel to the direction of the wave.

<h3>What is a longitudinal wave?</h3>

A longitudinal wave is a wave that is transversing along the length. When the displacement of medium and travel of wave is the same in that condition wave is known as the longitudinal wave.

It requires some medium to travel. A mechanical and sound wave is an example of a longitudinal wave.

Hence in the motion of the medium particles in a longitudinal wave, the medium vibrates parallel to the direction of the wave.

To learn more about the longitudinal wave refer to the link;

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yaroslaw [1]

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