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 The machine as shown has 2 rods of length 1 m connected by a pivot at the top.  The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has a roller that rolls along the floor in a slot.  As the roller goes back and forth, a 2 kg weight moves up and down.  If the roller is moving towards right at a constant speed, the weight moves up with a :

 

  • Option 1)

     constant speed

  • Option 2)

     decreasing speed

  • Option 3)

     increasing speed

     

  • Option 4)

    speed which is    \frac{3}{4}  th of that of the roller when the weight is 0.4 m above the ground

 

Answers (1)

 

From \Delta ABC \\ y = \sqrt{1- (\frac{x}{2})^2}

\frac{dy}{dt}= \frac{d}{dt}(\frac{\sqrt{4x^2}}{2}) = 1/2 \frac{d}{dx } (\sqrt { 4x^2})dx /dt

 

= \frac{1}{2}\cdot \frac{1}{2\sqrt{4- x^2}}(-2x )dx/dt

\frac{dy}{dt} = \frac{-x }{2 \sqrt {4 x^2 }}\cdot dx/dt

dx /dt = constant 

\Rightarrow \frac{dy}{dt} = \frac{-x }{2 \sqrt {4 x^2 }}\cdot dx/dt

\Rightarrow \frac{dy}{dt} = \frac{-x }{2 \sqrt {\frac{4}{x^2}-1 }}\cdot dx/dt

As with motion of roller , x decreases 

\therefore \sqrt{\frac{4}{x^2}-1 }\: \: increases

dy/dt decreases 

weight will move with decreasing speed 

 

 

 

 

 

 

 

 

 

 


Option 1)

 constant speed

Option 2)

 decreasing speed

Option 3)

 increasing speed

 

Option 4)

speed which is    \frac{3}{4}  th of that of the roller when the weight is 0.4 m above the ground

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subam

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