A man is driving around a curve rather quickly one day

  • Thread starter Thread starter HappyFlower
  • Start date Start date
  • Tags Tags
    Curve
AI Thread Summary
A man driving at 21.2 m/s around a curve has an object hanging from his mirror at a 36.4-degree angle from vertical, prompting a discussion on how to analyze the forces involved. Participants debated the approach to breaking down the forces acting on the object, specifically gravity and tension, and whether to separate them into components. The conversation highlighted the importance of considering centripetal acceleration alongside these forces. Different coordinate systems were discussed, with suggestions on which might simplify the problem-solving process. The discussion emphasizes the need for clarity in force analysis to determine the curve's radius effectively.
HappyFlower
Messages
30
Reaction score
1

Homework Statement


a man is driving around a curve rather quickly one day at 21.2m/s. He has an object hanging from his mirror that makes an angle of 36.4 wih respect to vertical as he goes around the curve. What is the radius of the curve?

Homework Equations

The Attempt at a Solution


I set up this but i don't know where to start.
Untitled.png
 

Attachments

  • Untitled.png
    Untitled.png
    3.7 KB · Views: 324
Physics news on Phys.org
You have written mg sin(θ) and mg cos(θ) on your diagram, but I cannot tell what you think these represent.
 
haruspex said:
You have written mg sin(θ) and mg cos(θ) on your diagram, but I cannot tell what you think these represent.
I was just making a diagram, don't know where to start.
 
HappyFlower said:
I was just making a diagram, don't know where to start.
I am not sure that I agree with your choice in separating gravity into components so soon. But let us see where it leads.

You've identified two forces (the components of gravity) and one acceleration on the object so far. Are there any other forces?
 
would i split up the tension instead that is in the string?
 
HappyFlower said:
would i split up the tension instead that is in the string?
If it were me, I'd likely have split tension into components. But let us see how far we can go having split up gravity instead (see note below).

Can you write down any equations relating mg cos(θ), mg sin(θ), tension and acceleration? (I am guessing that ∝ is intended to denote radial acceleration).

Note:
There are two forces and one acceleration on the object. Gravity, tension and centripetal acceleration. Gravity and centripetal acceleration are at right angles. If you decide to use a coordinate system at an angle to the one then it will be at an angle to the other as well. If you decide to use a coordinate system that is aligned with the one, then it will be aligned with the other as well. That's two simplifications. By contrast, if you align your coordinate system with the angle of the string you only get one simplification.

Either coordinate system will work, but I'm betting that one choice is easier than the other.​
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top