Calculating tension in rotational kinematics

AI Thread Summary
The discussion focuses on calculating the tension in a string as a block revolves in a circle with changing radius on a frictionless surface. The conservation of angular momentum is applied, leading to the relationship between the initial and final states of the system. The user derives the velocity as a function of the radius, concluding that v = (v1 * r1) / r. Subsequently, the tension in the string is expressed as T = (m * v1^2 * r1^2) / r^3. This formula effectively relates the tension to the initial velocity and the changing radius.
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Homework Statement


A block with mass m is revolving with linear speed v1 in a circle of radius r1 on a frictionless horizontal surface (see the figure ). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to r2.

Calculate the tension T in the string as a function of r, the distance of the block from the hole. Your answer will be in terms of the initial velocity v1 and the radius r1.

Homework Equations


I really have no idea. I'm going to assume it involves conservation of angular momentum: L = r X mv


The Attempt at a Solution


L1 = r1 * m * v1
L2 = r * m * v2

L1 = L2
r1 * m * v1 = r * m *v2
r1 * v1 = r * v2

Now I'm stuck

Thank you for the help
 
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Nevermind, using T = (mv^2) / r, and setting L1 = L2, I found v to be (v1*r1) / r

Plugged that into the tension and found T = (m*v1^2*r1^2)/r^3
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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