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Projectile on a string physics question

I am having some trouble with this question https://isaacphysics.org/questions/projectile_on_a_string?stage=all

So I think the time the tension on the rope is acting is t=v/g, using conservation of energy. So T=m*v^2/l, and from t=v/g the rope is loose, so it is projectile motion. Any help would be much appreciated
(edited 1 month ago)
Reply 1
Original post by Javier García
I am having some trouble with this question https://isaacphysics.org/questions/projectile_on_a_string?stage=all
So I think the time the tension on the rope is acting is t=v/g, using conservation of energy. So T=m*v^2/l, and from t=v/g the rope is loose, so it is projectile motion. Any help would be much appreciated

Not worked it through fully, but I would have thought that the angle when tension becomes zero would be important as that would give the initial conditions for projectile motion to pass through 0.
Original post by mqb2766
Not worked it through fully, but I would have thought that the angle when tension becomes zero would be important as that would give the initial conditions for projectile motion to pass through 0.

I am not sure about my working, but the angle would be Deltaphi=omega_0*t-1/2*alpha*t^2, where omega_0=v/l and alpha I believe is g/l? I know it does not have anything to do, but the ball is essentially going against gravity and the only "acceleration" it has is g. Hence alpha is g/l as the angular acceleration is the tangential acceleration divided by the radius. However I am not sure wether its g, I mean why would it be its tangential acceleration? What is the tangential acceleration that is making the ball decrease its velocity over time? Maybe g*sin(theta), as that is the one that is acting along the trayectory? Still, I am pretty sure it is not uniform circular motion, as the ball comes to a halt rotating about 0.
(edited 1 month ago)
Reply 3
Original post by Javier García
I am not sure about my working, but the angle would be Deltaphi=omega_0*t-1/2*alpha*t^2, where omega_0=v/l and alpha I believe is g/l? I know it does not have anything to do, but the ball is essentially going against gravity and the only "acceleration" it has is g. Hence alpha is g/l as the angular acceleration is the tangential acceleration divided by the radius. However I am not sure wether its g, I mean why would it be its tangential acceleration? What is the tangential acceleration that is making the ball decrease its velocity over time? Maybe g*sin(theta), as that is the one that is acting along the trayectory? Still, I am pretty sure it is not uniform circular motion, as the ball comes to a halt rotating about 0.

It would help to see a pic of your working. But write down the condition for circular motion (T>0) as a function of the current speed and relate that to the initial speed. You should get a relatively simple relationship between theta (angle when T becomes 0) and u (initial speed).
Original post by mqb2766
It would help to see a pic of your working. But write down the condition for circular motion (T>0) as a function of the current speed and relate that to the initial speed. You should get a relatively simple relationship between theta (angle when T becomes 0) and u (initial speed).

Sorry, but I am confused, the condition for circular motion is T=m*v^2/l, where v is the speed at any time, but how could I relate that to the angle?
Reply 5
Original post by javier garcía
Sorry, but I am confused, the condition for circular motion is T=m*v^2/l, where v is the speed at any time, but how could I relate that to the angle?

Its in the vertical plane so the speed will reduce and gravity will also act on the particle so its not just tension = centripetal. When the tension goes to zero, that will give the initial speed and the direction of motion is tangential to the circle so that will kick off the parabolic/projectile motion.
(edited 1 month ago)
I worked out the velocity when the tension is zero, and this is my working. I calculated the velocity of the ball when it reached point O, yet it is in terms of sin(theta). Help would be appreciated. PROJECTILE ON A STRING.jpg
Reply 7
Original post by javier garcía
I worked out the velocity when the tension is zero, and this is my working. I calculated the velocity of the ball when it reached point O, yet it is in terms of sin(theta). Help would be appreciated. PROJECTILE ON A STRING.jpg

Youve gone wrong at the start as the centripetal is not constant as the speed reduces due to gravity. So KE/GPE should get you there.

I cant see why the time is necessary, but once you have theta, that will give the initial speed and direction and required x-y displacement to plug into the parabolic x-y projectile equation. It would be worth writing out each step clearly and stating clearly what info is needed at each stage.
(edited 1 month ago)
Reply 8
I'm struggling with this problem too.

So far, I've got an expression for v in terms of sin(theta), which I derived from conservation of energy: v^2 = 3glsin(theta)

And I've got an expression for the velocity at the point where the string goes slack (I've called this v_f):

v_f^2 = glsin(theta).

But I'm going round in circles (no pun intended) trying to get a value or expression for sin(theta). Anyone had any luck?
Reply 9
Original post by ts316
I'm struggling with this problem too.
So far, I've got an expression for v in terms of sin(theta), which I derived from conservation of energy: v^2 = 3glsin(theta)
And I've got an expression for the velocity at the point where the string goes slack (I've called this v_f):
v_f^2 = glsin(theta).
But I'm going round in circles (no pun intended) trying to get a value or expression for sin(theta). Anyone had any luck?

the aim is to get the parabolic / projectile equation in terms of the initial speed v and youve done that as youve got the new initial speed and the angle it makes with the horizontal (trig value). So now its a case of making sure the parabolic equation passes through the origin.

Its easier to do if you transform the usual x-y coordinates and work along and perpendicular to the line where tension becomes zero. Though with hindsight the normal way isnt too bad if you make the right simplifications.
(edited 4 weeks ago)
Reply 10
Original post by mqb2766
the aim is to get the parabolic / projectile equation in terms of the initial speed v and youve done that as youve got the new initial speed and the angle it makes with the horizontal (trig value). So now its a case of making sure the parabolic equation passes through the origin.
Its easier to do if you transform the usual x-y coordinates and work along and perpendicular to the line where tension becomes zero. Though with hindsight the normal way isnt too bad if you make the right simplifications.

Unfortunately that's the bit I'm banging my head against a brick wall about.

Assuming the projectile starts from an initial height h = lsintheta), then the range of the projectile would be given by:

lcos(theta)=(v_f)cos(theta)t

or

l = sqrt(glsin(theta) t

But I'm a bit stumped with where to go next. I've now got an expression for sin(theta), but in terms of t.
Original post by ts316
Unfortunately that's the bit I'm banging my head against a brick wall about.
Assuming the projectile starts from an initial height h = lsintheta), then the range of the projectile would be given by:
lcos(theta)=(v_f)cos(theta)t
or
l = sqrt(glsin(theta) t
But I'm a bit stumped with where to go next. I've now got an expression for sin(theta), but in terms of t.

You want to put everything in terms of the initial speed v (and m, g and l) so your two relationships in #8 do that for sin(theta) and v_f (though you want to write sin(theta)=..., then sub that into the v_f=... ). That gives the initial conditions (speed and angle of projection) for the parabolic - projectile motion in terms of v,m,g and l and then using the projjectile equation(s) to determine the relationship between v and m, g and l which passes through the origin.

You can use pythagoras to get cos(theta) from sin(theta) and note that the initial angle of projection is perpendicular to the string when it becomes slack. So the initial direction of motion deals with the complementary angle. So you have got everything you need to set up the parabolic equations in terms of v, m, g and l by substituting the trig(theta) and v_f terms.

Also note the previous post. To grind the final relationship out, I found it easier to do the parabolic motion in terms of along and perpendicular to the string when tension becomes zero, though there is an extra bit of work to resolve gravity, rather than use the standard x-y coordinates. It obviously works for both coordinate systems through.
(edited 4 weeks ago)
Reply 12
Original post by mqb2766
You want to put everything in terms of the initial speed v (and m, g and l) so your two relationships in #8 do that for sin(theta) and v_f (though you want to write sin(theta)=..., then sub that into the v_f=... ). That gives the initial conditions (speed and angle of projection) for the parabolic - projectile motion in terms of v,m,g and l and then using the projjectile equation(s) to determine the relationship between v and m, g and l which passes through the origin.
You can use pythagoras to get cos(theta) from sin(theta) and note that the initial angle of projection is perpendicular to the string when it becomes slack. So the initial direction of motion deals with the complementary angle. So you have got everything you need to set up the parabolic equations in terms of v, m, g and l by substituting the trig(theta) and v_f terms.
Also note the previous post. To grind the final relationship out, I found it easier to do the parabolic motion in terms of along and perpendicular to the string when tension becomes zero, though there is an extra bit of work to resolve gravity, rather than use the standard x-y coordinates.

Thanks! I think I'm there now in terms of understanding what to do. I think my mistakes are mostly with the algebra now. Like you said, it seems to be a grind. I keep ending up with quadratics involving sin(theta)!
Original post by ts316
Thanks! I think I'm there now in terms of understanding what to do. I think my mistakes are mostly with the algebra now. Like you said, it seems to be a grind. I keep ending up with quadratics involving sin(theta)!

You can use (your #8)
sin(theta) = v^2/(3gl)
to eliminate sin(theta) from the equations as the terms on the right are what you want to be in your final equation.

If youre unsure, just upload your quadratics, but if done sensibly its not too bad, but its relatively easy to make it more complex than it needs to be. Note that the final relationship is in terms of v^4 so having sin^2 terms in there should be expected (but substitute as above).
(edited 3 weeks ago)
Reply 14
Original post by mqb2766
You can use (your #8)
sin(theta) = v^2/(3gl)
to eliminate sin(theta) from the equations as the terms on the right are what you want to be in your final equation.
If youre unsure, just upload your quadratics, but if done sensibly its not too bad, but its relatively easy to make it more complex than it needs to be. Note that the final relationship is in terms of v^4 so having sin^2 terms in there should be expected (but substitute as above).

Finally got it! Thanks for the help. That was one hell of an algebra exercise at the end.

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