I would like some help determining the expressions for the net forces acting on

Question

I would like some help determining the expressions for the net forces acting on the counter weight as well as the rubber stopper.

Lab 6 Centripetal Acceleration © John Griffith: April, 1999 Introduction In the last lab activity, you saw that a force applied along the direction of motion for an object caused the speed of that object to change but not the direction of travel. This week you look at what happens when a force is applied perpendicular to the direction of motion for an object. When this happens, the direction of the object changes, but its speed remains the same. In both of these cases, the velocity (vector) for the object is changing, which means that the object is accelerating. For the second case, where the object is changing direction but not speed, the acceleration is said to be a centripetal (or center-seeking) acceleration. The magnitude of this acceleration, ac, is related to the speed of the object, v, and the radius of the circular path, r, according to The direction of this acceleration is always radially inward, or toward the center of the circular path. For an object traveling at constant speed in a circular path, the net force acting on the object is also directed toward the center of the circular path, and the net acceleration the object experiences is given by the above expression: Since the object is moving at constant speed along this circular path, we can calculate the speed of the object if we know the period of motion, T, and the radius of the circular path, r distance 2r time For the portion of this lab dealing with centripetal force, you will determine an experimental value for the mass of your rubber stopper from the slope of a line. Procedure Centripetal Acceleration Start by attaching about 1.5 m of string to your rubber stopper. Pass the loose end of the string through the glass tube and then tie the safety swivel securely to the loose end. Measure a 1.0 m length of string from the center of the rubber stopper and place a piece of tape on the string for a reference marker. Place one 20 g mass on the swivel and lock the swivel closed. Before you start, make sure the area is clear for 1.5 m in every direction, as you are about to 25

Explanation / Answer

form the given diagram
let the mass of the cork be m
length of the string = L
angle of the rotation of the string be theta
then
from the diagram
r = Lsin(theta)

now, let tension in the stirng be T
then from force balance
Tcos(theta) = mg
Tsin(theta) = mv^2/r = mv^2/Lsin(theta)

tan(theta) = v^2/gLsin(theta)
sin(theta)tan(theta) = v^2/Lg

now time period of rotation be T'
then v = 2*pi*r/T' = 2*pi*Lsin(theta)/T'
v^2 = 4*pi^2*L^2*sin^2(theta)/T'^2

cos(theta) = 4*pi^2*L^2/T'^2*Lg
hence

for method 2
T'^2 = 4*pi^2*L/g*cos(theta) = constant

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