Centripetal Force and Acceleration

  • Thread starter kaotic
  • Start date
  • #1
kaotic
I have a problem here that looks simple to me, but has been driving me crazy:

Basically, it's an object weighing 3.33N at the end of a string of unknown length, which is in turn attached to a spring scale. The object is being spun in a circle with a radius of 1.49m. The person spinning the object is trying to spin the object with a velocity that will have the spring scale showing a weight of 4.44N (as opposed to the actual weight of 3.33N).

I need to determing the necessary velocity, the angle the object swings from vertical, the force the object exerts on the string, and the force necessary to keep the object moving on the circular path.

I tried using the equation Fc = mv^2/r, which I algebraically rearranged to solve for v, but I'm not sure if that's correct. If I put 4.44N in for Fc and solve for v, I get 4.4m/s, which doesn't seem right.

From there, I'm totally lost. If I could at least get the first two parts (v and angle), I'd probably be ok. I drew a free-body diagram to help me visualize the angle and figure it can be determine trigonometrically, but I don't think I have enough info to use sin and cos.

Any help you could give me to guide me on this would be appreciated.
 

Answers and Replies

  • #2
jamesrc
Science Advisor
Gold Member
476
1
Is the scale measuring the tension in the string? I'm a little unclear on where the scale is physically in the setup.

Assuming it is, though, you should be looking at your free body diagram. Let's call θ the angle that the string makes with the vertical so that the radial component of the tension is Tsinθ and the axial (vertical) component of the tension is Tcosθ.

In the vertical direction, that component of the tension balances with the weight of the object (3.3N) to give you one equation).

In the radial direction, that component of the tension balances the centripetal force, mv^2/r.

So we have two equations and two unknowns (θ and v), which you can solve for. (T, r, and m are known.)
 
  • #3
joc
26
0
draw the free body diagram - it's simply an object with 2 forces acting on it: tension and weight.

letting
T = tension in string,
m = mass of object,
w = weight of object = mg,
v = speed of object,
r = radius of rotation,
A = angle that string makes with vertical,

equating vertical forces:

TcosA = w --- (1)

and horizontal forces:

TsinA = mv^2/r --- (2)

now you can use eqn (1) to solve for A, then substitute it into eqn (2), then solve for v.

or you could be more indirect and apply the identity

sin^2(A) + cos^2(A) = 1 --- (3)

squaring eqns (1) and (2) and adding them up,

T^2.sin^2(A) + T^2.cos^2(A) = m^2.v^4/r^2 + w^2
T^2 [ sin^2(A) + cos^2(A) ] = m^2.v^4/r^2 + m^2.g^2

applying identity (3),

T^2 = m^2.v^4/r^2 + w^2
T^2 = (w/g)^2.v^4/r^2 + w^2

making v the subject,

v^4 = (gr)^2 [ (T/w)^2 - 1 ]

plugging in values,

v = 3.59ms^(-1)
 
  • #4
krab
Science Advisor
896
2
Originally posted by kaotic
I tried using the equation Fc = mv^2/r, which I algebraically rearranged to solve for v, but I'm not sure if that's correct. If I put 4.44N in for Fc and solve for v, I get 4.4m/s, which doesn't seem right...
That's your problem. Fc is not 4.44N, but Fc is the horizontal force while the weight, 3.33N is the vertical component. Together, they are [tex]4.44\mbox{N}=\sqrt{F_c^2+(3.33\mbox{N})^2}[/tex], because these two components are at right angles to each other. Solve this for Fc and then use your formula to get v. The rest is easy. For example, cosine of the angle is obviously 3.33/4.44=0.75.
 

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