My Math Forum Centripetal force

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 April 22nd, 2011, 11:12 PM #1 Joined: Dec 2010 Posts: 231 Thanks: 0 Centripetal force 1.A conical pendulum consists of a bob of mass 0.50kg attached to a string of length 1.0m.The bob rotates in a horizontal circle such that the angle the string makes with the vertical is 30° . Calculate a)the period of the motion b)the tension in the string 2.A simple pendulum is of length 0.5m and the bob has mass 0.25kg.Find the greatest value for the tension in the string when the pendulum is set in oscillation by drawing the bob to one side through an angle of 5.0° and releasing from rest.Explain where in the cycle the tension is greatest. i need help with these question.i don understand its concept and how to apply the formula.i need help and explanation for the question.Thank you so much.
 April 23rd, 2011, 11:29 PM #2 Global Moderator     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 11,556 Thanks: 101 Math Focus: The calculus Re: Centripetal force The Conical Pendulum A small body of mass m is suspended from a string of length L. The body revolves in a horizontal circle of radius r with constant speed v. Since the string sweeps out the surface of a cone, the system is known as a conical pendulum. We can find the speed of the body, the period of revolution $T_p$ and the tension T in the string. Drawing a free-body diagram for the mass m, where the force exerted by the string is T, we may resolve this force into a vertical component: $T\cos\theta$ and a component acting toward the center of rotation: $T\sin\theta$ Since the body does not accelerate in the vertical direction, the vertical component of T must balance the weight, thus: (1) $T\cos\theta=mg\:\therefore\:T=\frac{mg}{\cos\theta }$ Since the central force in this example is provided by the component $T\sin\theta$, from Newton's second law we get: (2) $T\sin\theta=ma_r=m\frac{v^2}{r}$ By dividing (2) by (1) we eliminate T and find: $\tan\theta=\frac{v^2}{rg}$ But, from the geometry, we note that $r=L\sin\theta$, thus: $v=\sqrt{rg\tan\theta}=\sqrt{Lg\sin\theta\tan\theta }$ Since the mass travels a distance of 2?r (the circumference of the circular path) in a time equal to the period of revolution, $T_p$ (not to be confused with the force T) we find: (3) $T_p=\frac{2\pi r}{v}=\frac{2\pi L\sin\theta}{\sqrt{Lg\sin\theta\tan\theta}}=2\pi\s qrt{\frac{L\cos\theta}{g}}$ 1.) m = 0.50 kg, L = 1.0 m, ? = 30°, g = 9.81 m/s² a) $T_p=2\pi\sqrt{\frac{$$1.0\text{ m}$$\cos$$30^{\circ}$$}{9.81\text{ \frac{m}{s^2}}}}\approx1.87\text{ s}$ b) $T=\frac{$$0.50\text{ kg}$$$$9.81\text{ \frac{m}{s^2}}$$}{\cos$$30^{\circ}$$}\approx5.66\t ext{ N}$ The Simple Pendulum A simple pendulum consists of a mass m attached to a light string of length L. The mass is released from rest when the string makes an angle $\theta_0$ with the vertical and the pivot is frictionless. If the mass is released from rest at the angle $\theta_0$, it will never swing above this position during its motion. At the start of the motion, position a, its energy is entirely potential. This initial potential energy is all transformed into kinetic energy at the lowest elevation, position b. As the mass continues to move along the arc, the energy again becomes entirely potential at position c, where $\theta=-\theta_0$. First, we need to find the speed of the mass at an arbitrary position d along the arc of motion, where $-\theta_0\le\theta\le\theta_0$. The only force that does work on m is the force of gravity, since the force of tension is always perpendicular to each element of the displacement and hence does no work. Since the force of gravity is a conservative force, the total mechanical energy is constant. Therefore, as the pendulum swings, there is a continuous transfer between potential and kinetic energy. If we measure the y coordinates from the center of rotation, then: $y_a=-L\cos\theta_0$ $y_d=-L\cos\theta$ Therefore: $U_a=-mgL\cos\theta_0$ $U_d=-mgL\cos\theta$ Applying the principle of constancy of mechanical energy gives: $K_a+U_a=K_d+U_d$ $0-mgL\cos\theta_0=\frac{1}{2}mv_d^2-mgL\cos\theta$ (1) $v_d=\sqrt{2gL$$\cos\theta-\cos\theta_0$$}$ We can see, without resorting to calculus, that $v_d$ is at its maximum value when $\theta=0$ Now we can find the tension in the string at point d. Since the force of tension does no work, it cannot be determined using the energy method. To find $T_d$, we can apply Newton's second law to the radial direction. First, recall that the centripetal acceleration is: $a_r=\frac{v^2}{r}$ directed toward the center of rotation. Since r = L, we get: (2) $\sum F_r=T_d-mg\cos\theta=ma_r=m\frac{v_d^2}{L}$ Substituting (1) into (2) gives for the tension at point d: $T_d-mg\cos\theta=m\frac{2gL$$\cos\theta-\cos\theta_0$$}{L}$ $T_d=mg\cos\theta+2mg$$\cos\theta-\cos\theta_0$$=mg$$3\cos\theta-2\cos\theta_0$$$ Again, it doesn't take calculus to see that the maximum tension occurs when $\theta=0$, or at the bottom of the swing, at position b. Thus, the maximum tension T is: $T=mg$$3-2\cos\theta_0$$$ Notice that if $\theta_0=0$ then T = mg, as we should expect. 2.) m = 0.25 kg, $\theta_0=5^{\circ}$ thus the maximum tension is given by: $T=$$0.25\text{ kg}$$$$9.81\text{ \frac{m}{s^2}}$$$$3-2\cos\(5^{\circ}$$\)\approx2.47\text{ N}$
 April 25th, 2011, 07:48 AM #3 Joined: Dec 2010 Posts: 231 Thanks: 0 Re: Centripetal force Thank you so much sir.Your explanation is excellent.I get the concept now.
 May 24th, 2011, 07:17 AM #4 Joined: May 2011 Posts: 23 Thanks: 0 Re: Centripetal force It is very useful to know roughly the answer BEFORE you start calculating it. In this case a conical pendulum is just (as always) a pendulum swinging two ways at once So its period (for a 30 degree swing) will be the same (only slightly less) than for the theory of "zero" swing Maybe you know the formula for that?

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