Question

Background/ Introduction 1. A small weight at the end of a string of length, L, forming a simple pendulum. 2. A horizontal rod is located a distance, d, directly beneath the pivot point. 3. The mass is held so that the string is taut and horizontal and then let fall. 4. What is the minimum distance, d, such that the mass will cause the string to loop over the pin at least In theory, we assume the bob is a point mass and the horizontal rod which is a distance d from the nfinitesimally thin. However, experimentally, we are looking for the minimum value of d once? such that the mass hits the lower rod and falls on the other side of it. The theoretical angle at which the string goes slack is The ratio of d/L is Some Physics principles to keep in mind: Since the only considered forces acting on the mass are tension and weight, and since the tension does no work on the mass (the tension is perpendicular to the Once the string goes slack, the tension disappears and the only force acting on the bob is the force of gravity. Thus, the mass is in projectile motion. displacement), energy of the mass is conserved. .At the point of the string going slack, the tension drops to zero, but at that moment there is still a centripetal force. Apparatus 2 Stands and Vertical Rods (at least one should be tall) At least 1-m of String 2 horizontal rods (at least one should be skinny) . 3 right angle connectors .Vernier Caliper 100g hooked mass (the bob) Lab Quest 2, 2 photogates . otes 1. "1-Step which does not have to be done in the lab. 2. Recorded measurements should be to the precision of the devices used. 3. Assumeg 9.796 m/s Position Meaning A--. Starting position. String is horizontal from the upper rod tothe mass. BA ertical from theupper rod to themass. C String is horizontal from the lower rod to the mass D String starts having slack E Mass just barely makes it over the lower rod. t bottom of arc. String is v -

Answer 1

When the mass is at a height of 'L', the potential energy is V = mg Now, when the reaches the lower position it has a kinetic energy = V Therefore 1/2 mv^2 = mg l Rightarrow v = squareroot 2gL Now, the make loop around the rod 'd' distance the pivot point, it will at least need this amount of K.E such that it can roll around the rod Therefore 1/2 m v^2 = 2 mg (L - d) Rightarrow gL = 2g (L - d) Rightarrow d = L/2 Rightarrow d/L = 1/2 we take the rest position of the bob of the Perpendicular to reference point than the angle at which the is block is180 degree