BR Physics, Ch2

[rebel1]

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Ex 2.4B

Which graph best represents the relationship between the cable tension and the magnitude of upward acceleration of an elevator?

I know there will be tension in the cable regardless of the motion of the elevator, so the grapgh has to be the one in which even when a=0, there is some tension. However, I don't understand how the tension and the acceleration would relate.

The correct answer is graph A, in which T increases linearly with 'a'



Another question I am having trouble with is,

Ex 2.9a

When a planet orbits the sun, a force acts upoon it that depends upon its orbital radius. How does the resulting tangential speed of the planet relate to the radius?

Answer: The speed is inversely proportional to the square root of the orbital radius.

Although TBR has provided an explanation for this one, but I still didn't understand it for some reason.
I was only usinf F=mv2/r and thought speed would be proportional to sq.rt of radius🙁

 

[Phantastic]

Another question I am having trouble with is,

Ex 2.9a

When a planet orbits the sun, a force acts upoon it that depends upon its orbital radius. How does the resulting tangential speed of the planet relate to the radius?

Answer: The speed is inversely proportional to the square root of the orbital radius.

Although TBR has provided an explanation for this one, but I still didn't understand it for some reason.
I was only usinf F=mv2/r and thought speed would be proportional to sq.rt of radius🙁

F=mv^(2)/r

so... v=sqrt(F*r/m)

but remember, F is still in there, and F decreases to a quarter of itself as you double the radius. So as you increase the radius, the orbital speed actually decreases, even though the relationship appears opposite at first. Add intuition.

 

[BerkReviewTeach]

Ex 2.4B

Which graph best represents the relationship between the cable tension and the magnitude of upward acceleration of an elevator?

I know there will be tension in the cable regardless of the motion of the elevator, so the grapgh has to be the one in which even when a=0, there is some tension. However, I don't understand how the tension and the acceleration would relate.

The correct answer is graph A, in which T increases linearly with 'a'

If you consider the elevator, there are two forces acting on it: mg in the downward direction and T in the cable in the upward direction. This means that the net force upward on the elevator can be found as ma = T - mg.

a will equal 0 when T = mg in magnitude, which means that the y-axis T has to have some positive value when a (the x-axis) is 0. As T increases, the value of ma will increase proportionally. Given that m is constant, this means that the increase in a is directly proportional to the increase in T, making for a line with an upward slope.

You could also rewrite the equation in the form of a line with T = ma + mg, where mg is the y-intercept and the slope is m.

 

[rebel1]

If you consider the elevator, there are two forces acting on it: mg in the downward direction and T in the cable in the upward direction. This means that the net force upward on the elevator can be found as ma = T - mg.

a will equal 0 when T = mg in magnitude, which means that the y-axis T has to have some positive value when a (the x-axis) is 0. As T increases, the value of ma will increase proportionally. Given that m is constant, this means that the increase in a is directly proportional to the increase in T, making for a line with an upward slope.

You could also rewrite the equation in the form of a line with T = ma + mg, where mg is the y-intercept and the slope is m.

Thank you so much for your help!

 

[rebel1]

F=mv^(2)/r

so... v=sqrt(F*r/m)

but remember, F is still in there, and F decreases to a quarter of itself as you double the radius. So as you increase the radius, the orbital speed actually decreases, even though the relationship appears opposite at first. Add intuition.

Thx Phantastic