Physics: force and motion

code PLUS

2012-04-11 10:46 pm

A student carries out an experiment to verify the centriputal force acting on a revolving object. A 250 g mass and a 1 kg mass are connected by a light string through a glass tube. The 250 g mass is then swung horizontally so that the length of the string that extends out of the tube remains constant at 0.5 m.

1. Can the upper part of the string be exactly horizontal?Why?
2. What is the tension in the string?
3. What is the net force acting on the 250 g mass?
4. The period of the 250 g mass is measured to be 0.69 s. Calculate the theoretical prediction of the centripetal force acting on it and the percentage error of the experiment.

回答 (1)

天同

2012-04-11 11:21 pm

✔ 最佳答案

1. No. If the string is exactly horizotnal, there is no force to support the weight of the 250 g mass. It would fall.

2. Since the 1 kg mass is in equilibrium, tension in the string equals its weight. Hence, string tension = 1g N = 10 N (if take g = 10 m/s2)

3. The 250 g mass is acted by two forces. The string tension and the weight of the mass. The weight is balanced by the vertical component of the string tension, leaving the horizontal component as the net force.

Hence, 10.sin(a) = (250/1000)g where a is the angle the string makes with the horizontal.
a = arc-sin[0.25] = 14.5 degrees
Net force = 10.cos(14.5) N = 9.68 N

4. Angular speed = 2.pi/0.69 s^-1 = 9.106 s^-2
Centripetal force = 0.25 x 0.5 x 9.106^2 N = 10.37 N

Percentage error = [(10.37 - 9.68)/9.68] x 100% = 7.1%

👍 0 👎 0