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Hi, I'm from Chile, so i had to translate the problem from spanish and my english ain't that good so if you don't understand something, please tell me.
A particle of mass m is initially at rest at the edge of a horizontal semicilinder of radius R (theta (t=0)=0). The particle is tied to an ideal string with the other end of the string (P) loose on the other side of the semicilinder. P is pulled vertically downwards, starting at rest and with a constant acceleration (a0). Depending on the value of a0 it is possible that in some point along the trayectory the particle detaches from the surfaceof the semicilinder or that the string looses its tension.
a)assuming the particle detaches from the surface before the string looses its tension, find an equation for the angle theta of detachment
b)assuming the string looses its tension before the particle detaches from the surface, find the angle theta in which this happens.
This is what I've got:
[tex]m(-\rho \dot{\theta^2} \hat{\rho} + \rho \ddot{\theta} \hat{\theta}) = N \hat{\rho} + T \hat{\theta} - mg\sin(\theta)\hat{\rho} - mg\cos(\theta)\hat{\theta}[/tex]
A particle of mass m is initially at rest at the edge of a horizontal semicilinder of radius R (theta (t=0)=0). The particle is tied to an ideal string with the other end of the string (P) loose on the other side of the semicilinder. P is pulled vertically downwards, starting at rest and with a constant acceleration (a0). Depending on the value of a0 it is possible that in some point along the trayectory the particle detaches from the surfaceof the semicilinder or that the string looses its tension.
a)assuming the particle detaches from the surface before the string looses its tension, find an equation for the angle theta of detachment
b)assuming the string looses its tension before the particle detaches from the surface, find the angle theta in which this happens.
This is what I've got:
[tex]m(-\rho \dot{\theta^2} \hat{\rho} + \rho \ddot{\theta} \hat{\theta}) = N \hat{\rho} + T \hat{\theta} - mg\sin(\theta)\hat{\rho} - mg\cos(\theta)\hat{\theta}[/tex]