Leaving Cert. Physics Question (Circular Motion)

oq4v3ht0u76kf2

I've been revising circular motion tonight and I've gotten most of the questions no problem, however this one has me stumped... I've attached an image of the setup of the system and the question is find the velocity of the particle at B.

g = 9.8 m/s

The string attached to the particle is a light inextensible string.

I keep coming up blank whenever I try this question. Any help is much appreciated.

For what it's worth, I tried using Theta = s / r and working out the length of the arc subtended but that's useless to me (I think) because velocity is obviously not constant or else the answer would just be 5 m/s

Sev

The easiest way to do this problem is to use the conservation of energy principle.

At the top of the arc, before you release the particle, it has initial velocity = Vi, at a height h. So its total energy is given by.

Ei = Potential Energy + Kinetic Energy
   = mgr + (1/2)mVi²

At the bottom of the flight, its height is 0, so the potential energy is 0 (mgr = mgr = 0) and its velocity is some unknown value Vf, that we want to solve for.

so, at the bottom of the arc

Ef = Potential Energy + Kinetic Energy
  = mg0 + (1/2)mVf²
  = (1/2)mVf²

Now Energy is conserved, so Ei = Ef.

    Ei = Ef
=> mgr + (1/2)mVi²  = (1/2)mVf²
=> 2gr + Vi² = Vf²    [cancelling m and multiplying across by 2]

Rearrange to get Vf on its own.

Vf = sqrt[2gr + Vi²]

Now plug in your values, g = 0.98m/s², Vi = 5m/s and r = 0.8m

And you get

Vf = sqrt[15.68 + 25]
    = sqrt[40.68]
    = 6.38m/s

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The alternative, newtonian, ie. F = ma method of solving this problem, the usual way a LC student would, is pretty nasty which is why I think its kind of a tough question for the LC student with F = ma drilled into them.

The other way goes like this.

Resolve the components of the force on the particle, mg, into two components. One acting perpendicular to its motion, ie, away from the string, and one acting in the direction of its motion, perpendicular to the string. The force perpendicular to its motion will always be cancelled out by the tension of the string. So the only force we need to worry about that affects the particles velocity is the component of gravity that acts perpendicular to the string.

If θ = 0 when the particle is released, and 90 at the bottom of its arc. Then this force can be written as.

F = mgcosθ = m(dv/dt)

=> gcosθ = dv/dt
         = d(rω)/dt (because v = ωr)
         = r(dω/dt) + ω(dr/dt)
         = r(dω/dt) (because dr/dt = 0)

so gcosθ = r(dω/dQ)

by the chain rule, dω/dt = (dθ/dt)(dω/dθ) = ω(dω/dθ)

=> gcosθ = rω(dω/dθ)
=> (g/r)cosθdθ = ωdω

The limits of Integration are then as follows.

For θ = 0, ω = Vi/r	(where Vi is the initial velocity, (v = ωr, therefore ω = v/r))
For θ = 90, ω = Vf/r	(where Vf is the final velocity)

Integrating then gives (g/r)Sinθ = ω²/2 + c

Put in the limits...

=> (g/r)[Sin(90) - Sin(0)] = [(Vf/r)² - (Vi/r)²]/2

so.. g/r = (1/2)(1/r²)(Vf² - Vi²)

Now we know everything except Vf, so rearrange for Vf.

Vf^2 = 2gr + Vi^2 

so Vf = sqrt[2gr + Vi²]

Which is exactly what we got last time.

Vf = sqrt[2(9.8)(0.8) + 5^2]
    = 6.38m/s

oq4v3ht0u76kf2

The thing is, the particle is being pushed from the bottom upwards... the final velocity solution in my textbook is 3.05 m/s. I probably should have put arrows on the diagram but my Paint skills are limited, the pink spot is the start position of the particle.

Sev

Oh shít.. oops.. I thought it was being released and falling from B to A

Ok.. well its not much different then. Same thing again

It has velocity Vi at the bottom of the arc, point A. It has kinetic energy and no potential energy.

Ei = Potential Energy + Kinetic Energy
   = mg0 + (1/2)mVi²
   = (1/2)mVi²

Then when it reaches point B.

Ef = Potential Energy + Kinetic Energy
  = mgr + (1/2)mVf²

Now Energy is conserved, so Ei = Ef.

    Ei = Ef
=> (1/2)mVi² = (1/2)mVf² + mgr
=> Vi² = Vf² + 2gr    [cancelling m and multiplying across by 2]

Rearrange to get Vf on its own.

Vf = sqrt[Vi² - 2gr]

Now plug in your values, g = 0.98m/s², Vi = 5m/s and r = 0.8m

And you get

Vf = sqrt[25 - 15.68]
    = sqrt[9.32]
    = 3.05m/s

---

To approach it the newtonian method, you'd do exactly the same thing as before, you'd just need to change the limits of integration. I dunno if you get all that integration stuff in the 2nd part, its just like the stuff you do in applied maths Q10 if you do it. I think all that calculus stuff tho its a bit too advanced for the LC physics course tho.

shark_eye

KE at A = KE at B + Potential energy at B

1/2 mv2 = 1/2 mv2 + mgh

1/2(0.24)(5)2=1/2mv2 + (0.24)(9.8)(0.8)

12.5 = 1/2v2 = 7.84

4.68 = 1/2v2

3.05 = v


centripetal acc = v2/r = 3.052/0.8 = 11.65

f= ma = 0.24(11.65)= 2.8 N


EASY pfffffffffff

ApeXaviour

shark_eye wrote: »

EASY pfffffffffff

And unnecessary, a three year old question that was answered well enough by Sev.

Please don't rehash old threads without some reason, it doesn't even have to be a good reason.