Question about Centripetal Force

[Banker]

Homework Statement

Question 2.b. from this paper - http://www.sqa.org.uk/files_ccc/PhysicsEQPAH.pdf

Homework Equations

let theta = x
Tcosx = mg
Tsinx = 2.5 (2.5 is the centripetal force)

The Attempt at a Solution

I rearranged the above equations and got 52 degrees as my answer. The mark scheme(at the bottom of the same document says it's 36 degrees.

 

[tommyxu3]

Maybe you can write down your way to find the angle?
You use the information you gave to get $\tan \theta?$

 

[Banker]

Maybe you can write down your way to find the angle?
You use the information you gave to get $\tan \theta?$

I did that and I got 52 degrees as my answer. The mark scheme says 36 but I'm not too sure if they used the correct value for mass.

 

[PeroK]

I did that and I got 52 degrees as my answer. The mark scheme says 36 but I'm not too sure if they used the correct value for mass.

You will need to show your working for people to help you properly. You haven't said at all how you use the angular velocity, for example.

 

[Banker]

You will need to show your working for people to help you properly. You haven't said at all how you use the angular velocity, for example.

F = mrw^2
F = 0.2(0.35)(6)^2
Central force = 2.5 N
Let Tension = T and the angle theta = x
From the diagram, Tcosx = mg
Tsinx = central force
hence Tsinx = 2.5
Dividing the first eq by the second:
Tsinx/Tcosx = 2.5/mg
Simplifying gives tanx = 2.5/mg
Weight, mg, = 0.2 x 9.8 (0.2kg is the mass of the plane)
hence tanx = 1.28
x = 52 degrees.

 

[PeroK]

F = mrw^2
F = 0.2(0.35)(6)^2
Central force = 2.5 N
Let Tension = T and the angle theta = x
From the diagram, Tcosx = mg
Tsinx = central force
hence Tsinx = 2.5
Dividing the first eq by the second:
Tsinx/Tcosx = 2.5/mg
Simplifying gives tanx = 2.5/mg
Weight, mg, = 0.2 x 9.8 (0.2kg is the mass of the plane)
hence tanx = 1.28
x = 52 degrees.

You're correct. You can see in the solution that they put $0.35$ for the mass. The funny thing is that the answer is independent of the mass of the plane! There was never any need to use the mass. I guess even the people who set these questions don't like algebra!

 

[Banker]

You're correct. You can see in the solution that they put $0.35$ for the mass. The funny thing is that the answer is independent of the mass of the plane! There was never any need to use the mass. I guess even the people who set these questions don't like algebra!

Ahh thank you for the clarification! Yeah the Scottish qualifications board are subpar compared their English counterparts.

 

[PeroK]

Ahh thank you for the clarification! Yeah the Scottish qualifications board are subpar compared their English counterparts.

I can't believe that!

 

[welovephysics]

I also got 52 degrees as my answer.

 

[haruspex]

You're correct. You can see in the solution that they put $0.35$ for the mass. The funny thing is that the answer is independent of the mass of the plane! There was never any need to use the mass. I guess even the people who set these questions don't like algebra!

It's worse than that. For part b they used the centripetal force calculated in part a. In part a, the correct mass was used, but in part b they got confused between the mass number, 0.2, and the radius number, 0.35. So in their calculation the masses did not cancel, which is why they got the wrong answer for the angle.
It's also a bit naughty that you have to pretend the plane develops no lift.