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Setler [38]
2 years ago
8

A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that

is perpendicular to the platform at its center. the angular speed of the merry-go-round is constant, and a child at a distance of 1.4 m from the axis has a tangential speed of 2.2 m/s. what is the tangential speed of another child, who is located at a distance of 2.1 m from the axis?
Physics
1 answer:
julsineya [31]2 years ago
3 0

Since angular speed is persistent, v/r is constant.
Where:

 v is tangential speed; and

 r is distance from axis.

 

Then equate v/r in both cases to get v in the second case.

Hence, speed = 2.2 x 2.1 / 1.4 meters

= 3.3 meters / seconds

 

Alternative solution would be:

w = 2.2 / 1.4 = 1.57

v = rw = 2.1 x 1.57 = 3.3 meters / seconds 

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Answer:

solution:

to find the speed of a jogger use the following relation:  

V

=

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x

/d

t

=

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×m

i

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in Above equation in x and t. Separating the variables and integrating,

∫

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×=

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=

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=

0

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now we have the relation to find the position and time for the jogger as:

−

4.7619  =

t

−

4.7619

.

.

.

.

.

.

.

.

.

(

2

)

Here

x  is measured in miles and  t  in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),    

     to get:

      = −

4.7619  

      =  

1

−

4.7619

      = −

3.7619

  or  x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in   m

i

l

/

differentiate  

     equation (1) with respect to time.

     we have to eliminate x from the equation (1) using equation (2).  

     Eliminating x we get:

     v

=

7.5×

     Now differentiating above equation w.r.t time we get:

      a

=

d

v/

d

t

       =

−

0.675

/

      At  

      t

=

0

      the joggers acceleration is :

       a

=

−

0.675

m

i

l

/

        =

−

4.34

×

f

t

/  

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        x

=

6  in equation (2).  We get:

        −

4.7619

(

1

−

(

0.04

×

6  )

)^

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/

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t

−

4.7619

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=

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h

r

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