Paper 4, Section I, B

Dynamics and Relativity | Part IA, 2013

A hot air balloon of mass MM is equipped with a bag of sand of mass m=m(t)m=m(t) which decreases in time as the sand is gradually released. In addition to gravity the balloon experiences a constant upwards buoyancy force TT and we neglect air resistance effects. Show that if v(t)v(t) is the upward speed of the balloon then

(M+m)dvdt=T(M+m)g.(M+m) \frac{d v}{d t}=T-(M+m) g .

Initially at t=0t=0 the mass of sand is m(0)=m0m(0)=m_{0} and the balloon is at rest in equilibrium. Subsequently the sand is released at a constant rate and is depleted in a time t0t_{0}. Show that the speed of the balloon at time t0t_{0} is

gt0((1+Mm0)ln(1+m0M)1)g t_{0}\left(\left(1+\frac{M}{m_{0}}\right) \ln \left(1+\frac{m_{0}}{M}\right)-1\right)

[You may use without proof the indefinite integral t/(At)dt=tAlnAt+C.\int t /(A-t) d t=-t-A \ln |A-t|+C . ]

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