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3 years ago

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An amusement park ride moves a rider at a constant speed of 14 meters per second in a horizontal circular path of radius 10. met

ers. What is the rider's centripetal acceleration in terms of g, the acceleration due to gravity?
A) 1g
B) 2g
C) 3g
D) 0g

1 answer:

ASHA 777 [7]3 years ago

4 0

Answer:

B) 2g

Explanation:

<u>Given the following data;</u>

Velocity, v = 14m/s

Radius, r = 10m

To find the centripetal acceleration;

$Acceleration, a = \frac {v^{2}}{r}$

Substituting into the equation, we have;

$Acceleration, a = \frac {14^{2}}{10}$

$Acceleration, a = \frac {196}{10}$

Acceleration, a = 19.6m/s²

In terms of acceleration due to gravity, g = 9.8m/s²

We would divide by g;

Acceleration, a = 19.6/9.8 = 2

Hence, centripetal acceleration = 2g

Therefore, the rider's centripetal acceleration in terms of g, the acceleration due to gravity is 2g.

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Explanation:

To analyze Which student is right it is best to propose the solution of the problem. Let's look for the speed of the center of mass. Let's use the concept of mechanical energy

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Em₀ = U = mg y

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Here the energy has part of translation and part of rotation

$E_{mf}$ = $K_{T}$ + $K_{R}$

$E_{mf}$ = ½ m $v_{cm}$ ² + ½ I w²

Where I is the moment of inertia of the body and w the angular velocity that relates to the velocity of the center of mass

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Let's replace

$E_{mf}$ = ½ I ( $v_{cm}$ / r)²

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This is the velocity of the center of mass of the bodies, as they all have the same radius with comparing this point is sufficient. Now let's use the speed definition

v = d / t

t = d / v

t = d / (√ [2gy / (1 + I / mr²)])

t = (d / √ 2gy) √(1 + I / m r²)

Therefore we see that time is proportional to the square root. All quantities are constant and the one that varies is the moment of inertia.

The moments of inertia of

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Cylinder Ic = ½ M r²

Hoop Ih = M r²

Let's replace each one and calculate the time

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ts = (d / √2gy) √ (1 + 2/5 Mr² / mr²)

ts = (d / √ 2gy) √ (1 +2/5) = (d / √ 2gy) √(1.4)

ts = (d / √ 2gy) 1.1

Cylinder

tc = (d / √2gy) √ (1 + 1/2 Mr² / Mr²)

tc = (d / √2gy) √ (1 + ½) = (d / √ 2gy) √ 1.5

tc = (d / √ 2gy) 1.2

Hoop

th = (d / √2gy) √ (1 + mr² / mr²)

th = (d / √2gy) √(1 + 1) = (d / √ 2gy) √ 2

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We have the results for the time the body that arrives the fastest is the sphere and the one that is the most hoop. Therefore the correct answer is

ts < tc < th

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3 years ago

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Answer:

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Explanation:

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What is the

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solution

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s = ut + 0.5×at² .............1

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Answer:

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Let us make L the subject of the formula:

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We are not told that the length of the string changes, hence, we can conclude that it is constant in both locations.

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$L = \frac{9.8 * 2^2}{4 \pi^{2}}\\ \\\\L = 0.9929 m$

When the pendulum is moved to a new location, the period becomes 1.99782 s.

We have concluded that length is constant, hence, we can find the new acceleration due to gravity, $g_n$ :

$0.9929 = \frac{g_n * 1.99782^2}{4\pi^{2}} \\\\\\0.9929 = 0.1011 g_n$

Therefore:

$g_n = 0.9929/0.1011\\\\\\g_n = 9.821 m/s^2$

The difference between the new acceleration due to gravity, $g_n$ and the former acceleration due to gravity, g, will be:

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3 years ago

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