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Rashid [163]
3 years ago
6

A playground merry-go-round with a radius of 2.0 m and a rotational inertia of 100 kg m2 is rotating at 3.0 rad/s. A child with

a mass of 22 kg jumps onto the edge of the merry-go-round, traveling radially inward. What is the new angular speed of the merry-go-round
Physics
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:

Explanation:

Conservation of angular momentum

The MGR originally has momentum

L = 100(3.0) = 300 kg•m²/s²

The child can be thought of as a point mass with I = mr²

When she jumps onto the rim of the MGR

300 = (100 + 22(2.0²)ω

ω = 300 / 188 = 1.5957... 1.6 rad/s

As she moves toward the center of the MGR, her moment of inertia goes to zero as her radius goes to zero.

The angular velocity when she reaches the center will again be 3.0 rad/s

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5 0
2 years ago
If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh
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Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

            v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

           a = v₀² / 2y

now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

7 0
3 years ago
a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cl
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Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m

Suppose, determine the time taken by the projectile to hit the ground.

We need to calculate the time

Using second equation of motion

s=ut+\dfrac{1}{2}gt^2

Where, s = vertical height of cliff

u = initial vertical velocity

g = acceleration due to gravity

Put the value in the equation

230=0+\dfrac{1}{2}\times9.8\times t^2

t=\sqrt{\dfrac{230\times2}{9.8}}

t=6.85 sec

Hence, The time taken by the projectile to hit the ground is 6.85 sec.

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