- #1
Nexus99
- 103
- 9
- Homework Statement
- Two blocks of mass M1 and M2 are hung as shown in the figure: ##M_1## is hung from the ceiling through an inextensible rope of negligible mass, while ##M_2## is hung on ##M_1## by an ideal spring of elastic constant k and rest length ##L_0##. At the initial moment (t = 0) the masses are stationary and the spring is compressed. For t = 0 the distance between ##M_1## and ##M_2## is equal to ##\frac{L_0}{2}##. In the next motion, ##M_2## goes down to reach a maximum distance ##L_{max}## from ##M_1##. Calculate
1. The magnitude of the tension T for t=0
2. The acceleration of ##M_2## for t = 0.
3. The value of Lmax and the corresponding magnitude of the tension T
4. The maximum speed reached by ##M_2## during the descent.
M1 = 1.20 kg, M2 = 0.25 kg, k = 30 N/m, L0 = 0.20 m.
- Relevant Equations
- conservation of energy, harmonic oscillator etc.
I tried in this way:
1) Considering a reference axis oriented downwards:
##M_1:## ## -T + M_1g - k \frac{L_0}{2} = 0##
##T = M_1g - k \frac{L_0}{2} ##
2)
##M_2:## ##M_2g + k \frac{L_0}{2} = M_2 a##
## a = g + \frac{k}{M_2} \frac{L_0}{2} ##
3) ## M_2g \frac{L_0}{2} + \frac{1}{2} k \frac{L_0^2}{4} = M_2gL_{max} + \frac{1}{2} k (L_{max} - L_0)^2 ##
## L_{max} = \frac{3}{2}L_0 - \frac{2gM_2}{k} ##
And
##T = M_1g + k(L_{max} - L_0) ##
4) Maximum velocity is reached when the spring is at equilibrium:
##M_2g \frac{L_0}{2} + \frac{1}{2}k (\frac{L_0}{2})^2 = M_2gL_0 + \frac{1}{2}M_2 v_{max}^2##
But solving this equation i got a complex value for ##v_{max}##:
##v_{max} = \sqrt{\frac{kL_0^2}{4M_2} - g L_0} ##
Where am i wrong?
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