A mass m is attached to a pole by two massless strings as shown in Figure 1, bel
ID: 1431424 • Letter: A
Question
A mass m is attached to a pole by two massless strings as shown in Figure 1, below. Both strings are of length L. The mass is circling the pole with a speed v that is high enough to keep the strings taut at an angle theta_1. The acceleration due to gravity g points downward on the page. The direction of rotation is indicated in the figures below You may ignore air resistance. Express all answer in terms of the variables m, L,g, theta_1. and theta_2. For Figure 1: Find the tensions (T_1 and T_2) for each string, at the instant shown. What is the magnitude and direction of the acceleration of the mass at the instant shown? For Figure 2: We now cut one string. The mass will be revolving around as a conical pendulum at an angle 02, as shown in Panel b. Calculate the velocity and period of rotation of the mass. If air resistance is not ignored, how will the velocity change? What will be the final state of the system? Explain your reasoning.Explanation / Answer
a) Force on m due to rotation = mv^2/r = mv^2/l*sin(theta-1)
Net force = mv^2/l*sin(theta-1) - mg = 2Tsin(theta-1) [T is the tensionn in the strin, same in both due to symmetry]
T = [mv^2/l*sin(theta-1) - mg]/[2sin(theta-1)]
Acceleration = v^2/l*sin(theta-1) - g
b) Let tension in string be T
Tcos(theta-2) = mg
Tsin(theta-2) = mv^2/r = mv^2/l*sin(theta-2)
tan(theta-2) = v^2/lg*sin(theta-2)
v = sqroot(lg*sin(theta-2)*tan(theta-2))
and period of rotation, T = 2*pi*l*sin(theta-2)/[sqroot(lg*sin(theta-2)*tan(theta-2))]
The velocity will decrease slowly, with the radius reducing in each uccesive rotation as the velocity slows down (because acc. is dependent on velocity)
Finally, when the radius is very small the motion might change into a pendulum and finally after a long time the system would come to rest eventually
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