# Tension in string

by pinkyjoshi65
Tags: string, tension
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 P: 263 A rock of mass 200g is attached to a 0.75m long string and swung in a vertical plane. A) what is the speed that the rock must travel so that its weight at the top of the swing is 0? B) what is the tension in teh string at the bottom of the swing? C) if the rock is now moved to a horizontal swinging position, what is the angle of the string to the horizontal? Here is what i could do: I drew a vertical circle with 2 positions a and b. A is at the top and B is at the bottom. so at A the F_g is down and T is up. at B,T is down and F_g is up. Right? then what?
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 Quote by pinkyjoshi65 i had another question.. A rock of mass 200g is attached to a 0.75m long string and swung in a vertical plane. A) what is the speed that the rock must travel so that its weight at the top of the swing is 0? B) what is the tension in teh string at the bottom of the swing? C) if the rock is now moved to a horizontal swinging position, what is the angle of the string to the horizontal? Here is what i could do: I drew a vertical circle with 2 positions a and b. A is at the top and B is at the bottom. so at A the F_g is down and T is up. at B,T is down and F_g is up. Right? then what?
 P: 9 This question requires you to use 2 concepts, i) that of the Newton' s laws of motion and ii)the conservation of energy. Firstly, if you've drawn a Free Body Diagram(FBD) correctly then you should get the tension $$T$$ upwards at the lowermost point in the trajectory(of the the rock), and downwards at the topmost point. Thats obvious because the string is inelastic (its length is always constant) so every little part of the string pulls the adjacent little part of the string with a force of the same magnitude (say $$dT$$) so that the string doesn't change in its physical dimension (to keep its length constant). Hence the direction of tension is always opposite to the force that is applied at its end, so that the length of the string remains same. Now moving specifically to the problem. Let the length of the string be $$l$$. We could assume the lowermost point to have a zero gravitational potential energy. Also let the velocity at the bottom of the loop be $$v_0$$ and at the top be $$v$$. First writing the energy conservation equation: $$\frac{1}{2} mv_0^{2} = \frac{1}{2}mv^{2} + 2mgl .... (1)$$ Where $$m$$ is the mass of the rock, $$g$$ is the acceleration due to gravity in the downward direction. This equates the initial kinetic energy of the rock at the bottom to the final mechanical energy of the rock(i.e. the kinetic energy plus the gained potential energy) at the topmost position. The second equation is the force equation at the top of the loop. Here we have: $$mg + T = \frac {mv^{2}}{l} .... (2)$$ Also since the the weight goes to zero at the topmost point, the tension should be minimum as the string is just about going to go slack, such that the centripetal force is equal to the weight. That is, we put $$T=0$$ in equation $$(2)$$; we also substitute the value of $$v$$ from the equation of conservation of energy to get the required answer ($$v_0$$). The answer should come to be $$v_0 = \sqrt{5gl}$$ That answers part a). For part b) balance the forces at the lowermost point. Here, the tension is upwards, the weight is downwards. $$T = mg + \frac {mv_{0}^{2} }{l}$$ Use the value obtained in a) to get the answer. Part c) the angle would obviously depend on the initial velocity of the rock when set into motion. It should behave like a conical pendulum. Balance the vertical components and horizontal components of the tension in string (I hope you get the direction right this time around) with the weight and the centripetal acceleration respectively. Use some trig. to get your answer. Part a) also gives the minimal velocity that has to be imparted to the rock at its lowermost position for it to complete one complete loop. Hope that helps!

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