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Consider this arrangement:

A point mass (its mass being $m$) is executing uniform circular motion on a frictionless horizontal plane. A string passing through the center of the circle and attached to the point mass is providing centripetal force. A force is being applied to the other end of the string such that circular motion is possible. The mass is moving with velocity $v_i$ within a circle of radius $R_i$. Let's now say the string gets pulled with even greater force, causing the radius of the trajectory to shrink to a value of $R_f$. The velocity of the mass point is now $v_f$.

The moment of tension (the only force applied to the mass point) with respect to the center $O$ of the circumference is null, as tension is always directed toward $O$. Therefore we can apply the conservation of angular momentum. The following relationship holds:

$$R_f V_f = R_iV_i = R(t)V(t)$$

From which:

$$V=\frac{R_i}{R}V_i$$

The point of the exercise is to show that the work done by tension equals the difference in kinetic energy between the time $i$ and the time $f$. This is generally true as stated by the work-energy theorem.

The exercise is solved in the following manner:

The difference in kinetic energy is :

$$K_f - K_i = \frac{1}{2}mV_f^2 - \frac{1}{2} mV_i ^2$$

$$\ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}m(V_f^2 - V_i^2)$$

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2} m \left(\frac{R_i^2}{R_f^2} V_i^2-V_i^2 \right)$$

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}mV_i^2 \left (\frac{R_i^2}{R_f^2} - 1\right)$$

No issue with that.

The infinitesimal work done by tension is $dW = -T(r)dr$ , where $T(r)$ is the tension when the radius of the trajectory is $r$, while $dr$ is the infinitesimal decrease of radius of the trajectory. Because tension acts at every moment as the centripetal force in a trajectory of radius $r$, and applying conservation of angular momentum, we say $T(r) = m\frac{V}{r^2} = m\frac{V_iR_i}{r^3}$. So the infinitesimal work done by tension when the trajectory's radius decreases by $dr$ is :

$$dW= -m \frac{V_iR_i}{r^3}dr$$

When we integrate from $R_f$ to $R_i$ (I'll skip the details) we get :

$$W= \frac{1}{2}mV_i^2 \left(\frac{R_i^2}{R_f^2} - 1\right)$$

which is exactly what we wanted.

My question(s):

I understand that considering tension to be equal to centripetal force is the only way to make the numbers work. What I'm failing to understand is how is it even possible that the radius of the circumference shrinks if the tension acts at every moment as the centripetal force for the trajectory of radius $r$ ? Shouldn't that keep the mass point in uniform circular motion ? How did the circumference start to shrink in the first place if a force never greater that the centripetal force was applied to the mass point ?

Also, when we say that the other end of the string gets pulled with more force than the centripetal force, aren't we saying exactly that ? Maybe tension in this particular case isn't constant along the string. It would be my first guess.

And what if we provided since the beginning a constant force which equaled the centripetal force needed for a circumference of radius $R_f$ ? Shouldn't we get the same result: the circumference shrinking until it stabilizes on that radius ? Clearly in the case of a constant force the work done by tension would be greater than that previously calculated.

On the other hand, the equality between work and difference in kinetic energy is a theorem, so the numbers must work. I'm quite confused.

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1 Answer 1

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In my opinion the best way to approach problems like this is using the Lagrange multiplier method. There is no potential, so the unconstrained Lagrangian is just the kinetic energy in polar coordinates: $$L=\frac{1}{2}m \left( \dot r^2 + r^2 \dot \theta^2 \right) $$ Now, the length of the string is controlled by some external force which enforces the constraint $r=R(t)$. This is added with a Lagrange multiplier to get the constrained Lagrangian: $$ L=\frac{1}{2}m \left( \dot r^2 + r^2 \dot \theta^2 \right) + \lambda \left( r-R \right) $$

Since the constrained Lagrangian does not depend on $\theta$ we get the conserved momentum conjugate to $\theta$ which is the ordinary angular momentum $$p_\theta = m r^2 \dot \theta = const.$$ Solving the Euler Lagrange equations for the constrained Lagrangian and using the constraint to eliminate $r$ gives $$\lambda = - m R \dot \theta^2+ m \ddot R = - \frac{p_\theta^2}{m R^3}+m \ddot R$$ $$\ddot \theta = -2\dot \theta \frac{\dot R}{R} = -2\frac{p_\theta}{m}\frac{\dot R}{R^3}$$ This gives us an expression for the tension in the string, $\lambda$, and the angular acceleration, $\ddot \theta$.

Therefore we can apply the conservation of angular momentum. The following relationship holds: RfVf=RiVi=R(t)V(t)

This is not quite correct. The conserved angular momentum is $p_\theta=mr^2\dot \theta$, but because in general $\dot r \ne 0$ we cannot say in general that $r\dot \theta = v$.

Because tension acts at every moment as the centripetal force in a trajectory of radius r

In the Lagrange multiplier method, the tension is $\lambda$. So, we can see that it is not correct to assume that the tension is simply the centripetal force. The first term in $\lambda$ is indeed the centripetal force, but there is also a term $m\ddot R$. This makes sense, if you consider the case where the conserved angular momentum is 0 then the centripetal force term is 0 but the tension is not. The tension in that case would be entirely due to the $m \ddot R$ term.

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