Physics, 23.09.2019 08:20 dexterwilliams161

A44.0-kg child takes a ride on a ferris wheel that rotates four times each minute and has a diameter of 20.0 m.
(a) what is the centripetal acceleration of the child?
(b) what force (magnitude and direction) does the seat exert on the child at the lowest point of the ride?
(c) what force does the seat exert on the child at the highest point of the ride?
(d) what force does the seat exert on the child when the child is halfway between the top and bottom? (assume the ferris wheel is rotating clockwise and the child is moving upward.)

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Answer from: cranfordjacori

(a) 1.76 m/s^2

The centripetal acceleration of the child is given by:

$a_c=\omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the wheel

The radius of the wheel is half the diameter:

$r=\frac{d}{2}=\frac{20.0 m}{2}=10.0 m$

The wheel makes 4 revolution per minute, so the angular velocity is

$\omega=4 rev/min$

Let's remind that

$1 rev = 2 \pi rad$

1 min = 60 s

So the angular velocity is

$\omega=(4 rev/min) \cdot \frac{2 \pi rad/rev}{60 s/min}=0.42 rad/s$

So, the centripetal acceleration is

a_c=(0.42 rad/s)^2(10.0 m)=1.76 m/s^2

(b) 509.1 N, upward

At the lowest point of the ride, we have the following forces:

- Normal force exerted by the seat on the child: N, upward

- Weight of the child: W = mg, downward

The resultant of these forces must be equal to the centripetal force, which is upward (towards the centre of the wheel), so we have the following equation

N-mg = ma_c

From which we can find the normal reaction of the seat on the child:

N=m(g+a_c)=(44.0 kg)(9.81 m/s^2+1.76 m/s^2)=509.1 N

At the highest point of the ride, we have the following forces:

- Normal force exerted by the seat on the child: N, upward

- Weight of the child: W = mg, downward

The resultant of these forces must be equal to the centripetal force, which this time is downward (towards the centre of the wheel), so we have the following equation

mg-N = ma_c

From which we can find the normal reaction of the seat on the child:

N=m(g-a_c)=(44.0 kg)(9.81 m/s^2-1.76 m/s^2)=354.2 N

(d) 431.6 N, upward

When the child is halfway between the top and the bottom, the normal force exerted by the seat on the child is simply equal to the weight of the child; therefore we have:

N=mg=(44.0 kg)(9.81 m/s^2)=431.6 N

Answer from: Quest

distance on the y-axis and time on the x-axis.

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the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

slope of a graph = change in value of y -axis / change in values of x -axis

comparing both the equations

change in value of y -axis = change in position

change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

Answer from: Quest

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A44.0-kg child takes a ride on a ferris wheel that rotates four times each minute and has a diameter...

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