A 0.3-kg object is being whirled in a horizontal circle at the end of a 1.5 m long string. If the string breaks when the number of revolutions per minute, rpm = 200, then find the maximum tension in the string. A) 2.0  10² N B) 59 N C) 7.0  10⁵ N D) 9.0 N E) 88 N

Question

Quizplus · Accepted Answer

Explanation:The maximum tension in the string occurs when the object is at the top of its circular path. At this point, the tension in the string is equal to the sum of the gravitational force and the centripetal force. The gravitational force is given by:$$F_g = mg = (0.3 kg)(9.8 m/s^2) = 2.94 N$$The centripetal force is given by:$$F_c = mv^2/r$$where m is the mass of the object, v is its speed, and r is the radius of the circle. The speed of the object is given by:$$v = 2πrf$$where f is the frequency of revolution. The frequency of revolution is given by:$$f = 200 rpm = 200/60 Hz = 3.33 Hz$$The radius of the circle is given by:$$r = 1.5 m$$Substituting these values into the equation for centripetal force, we get:$$F_c = (0.3 kg)(3.33 Hz)^2(1.5 m) = 3.56 N$$The maximum tension in the string is given by:$$T = F_g + F_c = 2.94 N + 3.56 N = 6.5 N$$Therefore, the correct answer is A.

A 03-Kg Object Is Being Whirled in a Horizontal Circle at at the End