What is the Speed of the Train as it Rounds a Curve?

[julz3216]

Homework Statement

A train traveling at a constant speed rounds a curve of radius 218 m. A lamp suspended from the ceiling swings out to an angle of 16.6° throughout the curve. What is the speed of the train?

Homework Equations

mv^2/r

The Attempt at a Solution

I drew a diagram and attempted to calculate v by setting mv^2/r = -cos16.6 mg
I don't really know how to approach this?

 

[LowlyPion]

Homework Statement

A train traveling at a constant speed rounds a curve of radius 218 m. A lamp suspended from the ceiling swings out to an angle of 16.6° throughout the curve. What is the speed of the train?

Homework Equations

mv^2/r

The Attempt at a Solution

I drew a diagram and attempted to calculate v by setting mv^2/r = -cos16.6 mg
I don't really know how to approach this?

I think the angle is with the vertical, which makes the deflection in x given as Sin16.6 not Cos 16.6.

V²/r = Sin16.6*g

V = (r*sin16.6*g)^1/2

 

[julz3216]

I tried and got 24.705 but that was wrong. Are there any other ways to approach the problem?

 

[LowlyPion]

I tried and got 24.705 but that was wrong. Are there any other ways to approach the problem?

What units do they want the answer in? m/s or km/h?

24.705 m/s = 88.9 km/h

 

[julz3216]

I tried both ways but neither options were correct. I think it is supposed to be in m/s and I think my answer is wrong in general, is there anything else I can do to get another answer?

 

[LowlyPion]

Ooops. Sorry. I did a sketch and realized vertical is g and that means then that

V² = tan16.6*g*r

 

[julz3216]

Ok, I got it! Thank you so much.

 

[LowlyPion]

Ok, I got it! Thank you so much.

It's important you understand why.

Draw the acceleration vectors. The acceleration vectors add to some resultant a that forms the angle. The vertical component is g which means the Resultant acceleration on the lamp is given by

g = a_y = a*cosθ

So a = g/cosθ

For the x component, that means that a_x = a*sinθ = g*sinθ/cosθ = g*tanθ

and that is what equals the centripetal acceleration.