# Variable Mass:Sand from balloon

1. Apr 17, 2013

### CAF123

1. The problem statement, all variables and given/known data
A balloon of mass M contains a bag of sand of initial mass $m_o$. The balloon experiences a constant upthrust and is initially at rest and in equilibrium: the upthrust compensates exactly for the gravitational force. At time t=0, the sand is released at a constant rate. It is fully disposed of in time $t_o$. Determine the velocity $v(t)$ at any time between t=0 and $t=t_o$. Express the result in terms of $$k = \frac{m_o}{M+m_o} \frac{1}{t_o}$$

3. The attempt at a solution

At t=0, sand released at constant rate, $\alpha$ say. So, $\frac{dm}{dt} = -\alpha,$m the mass of the sand bag. Solve the above to give $m_s(t) = m_o - \alpha t$

The mass of the whole system (balloon + sand bag), $M(t) = M + m_o -\alpha t$

By NII, $$P(t+\Delta t) - P(t) = F_{ext} \Delta t \Rightarrow M(t + \Delta t)v(t + \Delta t) - M(t)v(t) = F_{ext}\Delta t$$

Taylor expand the first two terms to give $$(M(t) + \Delta t \dot{M})(v(t) + \Delta t \dot{v}) - M(t)v(t) = [(M + m_o)g - M(t)g]\Delta t,\,\,F_{ext} = (M + m_o)g - M(t)g$$

Collect terms ignore the (very small) $\Delta t^2$ term to get $$\dot{v} - \frac{\alpha}{M + m_o - \alpha t}v = \frac{(M+m_o)g - M(t)g}{M + m_o - \alpha t},$$ which can be solved by integrating factor.

Does it look good so far? When I look at the solution, they end up doing just an ordinary integration.

Many thanks.

Last edited: Apr 17, 2013
2. Apr 17, 2013

### TSny

$P(t+\Delta t)$ should include the momentum of all of the material that is included in $P(t)$. That would include not only the material still left in the balloon at $t+\Delta t$ but also the sand material that was released during the interval $\Delta t$.

3. Apr 17, 2013

### CAF123

$P(t + \Delta t)$ is the momentum of the material still left in the balloon after $\Delta t$ and the momentum of the sand released in that $\Delta t$. So I should have had something like: $$P(t + \Delta t) = M(t+\Delta t)v(t+\Delta t) - dm(v+dv),$$ dv the change in velocity as a result of disposing the sand.

This will reduce the eqn to $$M(t)dv = (M+m_o)g - M(t)g$$ Is it better?

4. Apr 17, 2013

### TSny

Yes, that looks good (but you dropped a $dt$ somewhere).

5. Apr 17, 2013

### TSny

Maybe a minor point: For the last term $dm(v+dv)$ you could use just $v$ instead of $v+dv$ since you only need terms to first order.

6. Apr 17, 2013

### CAF123

So integrating from v(t=0) to v and from t=0 to t gives: $$\int_0^v dv = (M+m_o)g\int_0^t \frac{dt}{M+m_o - \alpha t} - gt$$

This gives $$v(t) = \frac{(M+m_o)g}{\alpha} \ln \left(\frac{M+m_o}{M+m_o-\alpha t}\right) - gt$$

Is there another way to express $\alpha$?

7. Apr 17, 2013

### TSny

You should be able to express $\alpha$ in terms of $m_0$ and $t_0$.

8. Apr 17, 2013

### CAF123

That would be $\alpha = m_o/t_o$.

9. Apr 17, 2013

Right.