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Alexxandr [17]
3 years ago
13

A playground merry-go-round has radius 2.10m and moment of inertia 2500kg*m^2 about a vertical axle through its center, and it t

urns with negligible friction.
A)A child applies an 18.0 N force tangentially to the edge of the merry-go-round for 16.0s . If the merry-go-round is initially at rest, what is its angular speed after this 16.0s interval?


in rad/s

B)How much work did the child do on the merry-go-round?



C)What is the average power supplied by the child?


I understand the jist of the question but am stuck on where to start
Physics
1 answer:
vladimir2022 [97]3 years ago
8 0

Answer

Radius of the wheel r = 2.1 m

Moment of inertia I = 2500 Kg m²

Tangential force applied F = 18 N

Time interval t = 16 s

Initial angular speed ω1 = 0

Final angular speed ω2 = ?

Let α be the angular acceleration.

Torque applied τ = Iα

                         F r = Iα

Angular acceleration α = F r/I

                                    = \dfrac{18\times 2.1}{2500}

                                    = 0.015 rad/s²

(a)From rotational kinematic relation

            Final angular speed ω₂ = ω₁ + αt

                                                 = 0 + (0.015 rad/s^2 * 16 s)

                                                 = 0.24 rad/s

(b) Work done W = 0.5 Iω₂² - (1/2)Iω₁²

                      = 0.5*( 2500 Kg m²)(0.24 rad/s)^2 - 0

                      =  72 J

(c) Average power supplied by the child P = W/t = \dfrac{72}{16}

                                                                        = 4.5 watt        

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Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
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Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

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Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

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Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

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v = (21760/4800)

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So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

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Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

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F × 0.110 = 895.3

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