# Homework Help: Vertical circular motion of a body

1. Aug 5, 2009

### leena19

1. The problem statement, all variables and given/known data
A mass of 100g is attached to a string of length 1m.The system experiences vertical circular motion.
(1)What is the minimum velocity to be given to the object at the lowest point of the path, in order to continue vertical circular motion?

(2)What is the maximum tension in the string?

(3)What is the tension in the string when the string makes an angle of 60 degrees with the vertical?

2. Relevant equations

F=ma
and equations of motion maybe

3. The attempt at a solution

I have done problems on conical pendulums,but never on vertical circular motion,so I'm not sure how I'm supposed to find the answer to (1)

Anyway at the lowest point,I know
T = mg + mv2/r
And I can find v,if I knew the tension T ?but anyway this v is only the centripetal velocity,right?
Not sure if knowing v would even help me find the minimum velocity(u) at the lowest point?

Hope my problems clear
THANK YOU.

2. Aug 5, 2009

### Staff: Mentor

Do a similar analysis for the highest point. Hint: What value of tension will allow you to solve for the minimum speed at the highest point?

3. Aug 5, 2009

### leena19

T+mg =mv'2/r

When T=0,
mg=mv'2/r ?

or can we take v'=0
then T=-mg ?which in this case means the mass would drop vertically down,through the midpoint?

4. Aug 5, 2009

### Staff: Mentor

Good.

No. The minimum speed is the speed where the string just begins to go slack, thus where T = 0. That minimum speed is greater than zero! (Below that minimum speed and the mass will not be able to follow a circular path.)

5. Aug 5, 2009

### leena19

Just curious.How would it move below this speed?Would it be a projectile-like motion?
And if it were to freefall right through the centre of its motion,then both T and v would have to be 0?

Now back to the original question,
At the highest point,
mg=mv'2/r
v'2= 10

Then is it OK to assume the object moves with uniform velocity throughout and so take the minimum velocity at the lowest point also as sqrt(10) ?
Or do I assume the velocities are different and use conservation of energy to solve the whole problem?

Sorry, I really don't know what to do.

6. Aug 5, 2009

### RoyalCat

Yes, that is correct.
If the mass begins with less than the minimum velocity, then it will only reach a certain height before its string goes slack, at which point it will enter projectile motion according to the direction of its velocity at the moment the tension turned to 0.

Your analysis was correct, for the minimum velocity for the object to remain in circular motion throughout the rotation, then we demand that the tension only turn to 0 at the highest point. Note that as the object descends, its velocity increases, which in turn increases the $$\tfrac{mv^2}{r}$$ term and produces a non-zero tension in the string.

Energy is conserved in this system. That is to say, $$U_g+E_k=constant$$
As you know the velocity at the top of the rotation, and the height the mass descends in its rotation, you can find its velocity at the bottom of the rotation just as well.

You would be completely wrong to assume that this is circular motion with uniform velocity. Gravity increases the velocity of the mass as it descends, and decreases it as it ascends (If you're comfortable with energy, the kinetic energy turns into potential gravitational energy ("Height energy") and vice-versa).

7. Aug 5, 2009

### leena19

Thanks so much for the very detailed and clear explanation!

For part (1), applying the law of conservation of energy and taking the lowest point as the reference level,
1/2mv2 = mgh + 1/2mv'2 where v is the velocity of the body at the lowest point.

v2 = 2gh + v'2
v2 = 2*10*2 + 10
v = sqrt 50
=7.07m/s ?

Now for part (2),
Tmax = mg + mv2/r
= 1 + (0.1*50)
=6N?

part(3)

The question asks us to find the tension in the string when it makes an angle of 60 degrees with the vertical,but I don't know which one to take?
the lower one,i.e. when
T1 = mgcos60 + mv"2/r ?
or the one positioned higher with respect to the lowest point of its circular motion,i.e
T1 + mgcos60 = mv"2/*r ?

8. Aug 5, 2009

### Staff: Mentor

All good.

The question is ambiguous. Do both.

9. Aug 5, 2009

### leena19

Applying coservation of energy to the object when it is at the lower point,

1/2mv2 = mgh + 1/2mv"2 where h=0.5m
v2= 5+ v"2
50-5=v"2
v"2=45

Substituting this in the tension equation
T1 = mgcos60 + mv"2/r gives
T1 = 0.5+(0.1*45)
T1 = 5N

Then doing the same for the other point,
1/2mv2 = mgh + 1/2mv"2 where h is now 1.5m
v"2=35
T1 + mgcos60 = mv"2/r
T1=3N

I haven't got the answers to check,but I'm fairly confident they are correct,cause i feel I've understood the problem very well.
All thanks to you and RoyalCat!
Thank you very much for all the help!