Changing Mass

Classical Mechanics Level 4

In the above pulley-mass system the mass m m constantly decreases with a rate of 0.01 g s − 1 0.01gs^{-1} .

If initially both the masses m , M m,M were equal to 1 K g 1Kg , then find the distance travelled by the mass M M in meters in 30 s 30s assuming that the mass m m didn't reached the top end in those 30 s 30s and a c c e l e r a t i o n d u e t o g r a v i t y o f e a r t h ≈ 10 m s − 2 \red{acceleration\space due\space to\space gravity\space of\space earth}\approx 10ms^{-2} .


The answer is 0.225.

Zakir Husain by Zakir Husain , 41, India

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1 solution

Riccardo Baldini
Jun 8, 2021

Since the mass m m decreases at constant rate α = 0.01 g s − 1 \alpha = 0.01gs^{-1} and at the beginning m ( 0 ) = M m(0)=M , we can write:

m ( t ) = M − α t m(t)=M-\alpha t

The force on the system is:

F = ( M − m ( t ) ) g = ( M + m ( t ) ) x ¨ ⇒ x ¨ = ( M − m ( t ) ) g M + m ( t ) F=(M-m(t))g=(M+m(t))\ddot{x}\Rightarrow \ddot{x}=\frac{(M-m(t))g}{M+m(t)}

x ¨ = α g t 2 M − α t \ddot{x}=\frac{\alpha gt}{2M-\alpha t} , with boundary conditions x ( 0 ) = 0 and x ˙ ( 0 ) = 0 x(0)=0\text{ and }\dot{x}(0)=0

It's easy to integrate that expression twice in order to get x ( t ) x(t) :

x ( t ) = α t ( 4 M − α t ) + 4 M log ⁡ ( 2 M ) ( α t − 2 M ) + 4 M ( 2 M − α t ) log ⁡ ( 2 M − α t ) 2 α 2 g x(t)= \frac{\alpha t (4 M-\alpha t)+4 M \log (2 M) (\alpha t-2 M)+4 M (2 M-\alpha t) \log (2 M-\alpha t)}{2 \alpha ^2} g

Using the values provided in the text, we fine x ( 30 s ) = 0.225 m x(30s)=0.225m .

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