Finding the measurements for F, M, r, and f using circular motion apparatusBhavi Patel Physics 131 Lab section 30 Lab Room 210 10/24/12 Abstract:In this experiment, the goal was to obtain the measurements for F, f, M, and r to verify thefollowing equation F = 4π2f2Mr. We did this by using a circular motion apparatus. First we started off by measuring the force on the spring carrying M as it rotated, then we measured r by measuring the area between the pointer and the rotatable vertical shaft, we measured F by addingweights on the hanger attached to the string and pulley until M was hanging directly above the pointer, and to find f we removed the string and hanger and twisted the vertical shaft until M passed over the pointer and using the timer to calculate the frequency. We then repeated this procedure 2 more time with different r and recorded our data. Questions:1.) An object rotating on the circumference of a circle is known as circular motion.a. An object in uniform circular motion is accelerating due its change in direction.b. Centripetal acceleration is the rate of change in velocity at which in object is moving in the orbital, it is directed towards the center of rotation. Along the radiusthe vector of the circular motion is inwards for the direction and the magnitude is related to the angular velocity and speed. 2.) Doubling the speed of the car would have a greater effect on the magnitude of its acceleration. This is because when you double the speed of the car the acceleration becomes quadrupled but if you move to a track with half the radius the acceleration only doubles. We would calculate this by using the formula for centripetal acceleration A = V² / R, I would then compare the final acceleration with the initial acceleration Af = (Vf) ² / R Ai = (Vi) ² / R Double Velocity:Af = (2 × Vi)² / R To determine the ratio of the final acceleration to the initial divide the two: = [(2 × Vi)² / R] / [(Vi)² / R] = 4 × Vi² / Vi² = 4Radius is halved:Rf = Ri/2. To determine the ratio of the final acceleration to the initial divide the two: Af / Ai = [V² / (Ri ÷ 2)] / [V² / (Ri)] = 23.) Centripetal force is centripetal force is defined as a force which keeps a body moving with a uniform speed along a circular path and is directed along the radius towards the centera. In this experiment centripetal force is provided by the rotating vertical shaft and spring. b. When you go around a corner in a car, you feel centrifugal force.4.) Explain why ∣F=mg∣ = ∣Fc=4 π2M f 2 r∣ in our experiment.5.)r(meters)m (kg) F=mg (N) f sec^-1 4π2f2Mr (N)0.215 0.655 6.42555 1.46483 8.221370.195 0.485 4.75785 1.25471 5.444030.175 0.37 0.6297 1.21212 4.559626.)a.) The expected slope of the line graph is a half because when F=mg vs. F = 4π2f2Mr when they are equaled one is 2 and the other is a half. b.) The actual line of best fit for our data was 0.5524x + 3.8999. The percent of error for the slope is 1.367. The source of error is that the velocity was not constant because we were rotating the shaft. 7.) I would expect the y-intercept for my line of best fit to be 3.89.To find the y-intercept you have to make x=0,Y = 0.5524(0) + 3.899Y= 3.8998.) If a harder spring was used during the experiment, this would cause the centripetal force Fc to increase because more m would need to be added to bring the pointer to the center and it would cause the period f of the hanging mass to increase because the harder spring would cause the momentum to increase.Conclusion In this experiment, we measured the quantities F, f, M, and r to verify the following equation F = 4π2f2Mr. With these values we were able to compare the spring force on the left hand side (F = mg) with the centripetal force on the right hand side (4π2f2Mr) and we saw that they are related. After comparing the two equations, we realized that one is doubled and one is a half, therefore we assumed that the slope should be around .5 and later when we found the line of best fit we were able to determine this because we got a slope of