A yo-yo of mass M 0.130 kg and outer radius R that is 4.25 times greater than th

Question

A yo-yo of mass M 0.130 kg and outer radius R that is 4.25 times greater than the radius of its axle r is in equilibrium if a mass m is suspended from its outer edge as shown in the figure below Determine the tension in the two strings, T1 and T2 and the mass m. 0.04 Since the yo-yo is in static equilibrium, we expect to use the first and second conditions of equilibrium. Also since we have three unknowns, we should expect to develop three equations which may Ti be solved simultaneously to obtain the three unknowns. Since m is in translational equilibrium, see if you can write a first condition of equilibrium statement that relates m to T2. Since the yo-yo is in rotational equilibrium, see if you can write a second condition of equilibrium statement that relates T1 and T2. Finally, since the yo-yo is also in translational equilibrium, see if you can write a first condition of equilibrium statement that relates M, T, and T2. These three statements amount to three equations in terms of the three unknowns. See if you can manipulate these three equations in order to determine the tension T1. N 1.666 T2 Now that we know the tension T1 from the first response to the problem, we have two options for obtaining the tension T2. One option uses the first condition of equilibrium and the other uses the = second condition of equilibrium. See if you can write a first condition of equilibrium statement for the vertical forces that will allow you to determine the tension T2, now that we know the tension T1 As a check on your work, see if you can write a second condition of equilibrium that will allow you to determine the tension T2, now that we know the tension T1. N 0.392 Now that we know the tension T2 from the second response to the problem, see if you can write a first condition of equilibrium statement for the vertical forces that will allow you to determine the mass m. kg r21

Explanation / Answer

Let:
T1 be the tension at the axle (not labelled in the diagram),
T2 be the tension at the edge (not labelled in the diagram),
r be the axle radius,
R be the outer radius,
M be the mass of the yoyo,
m be the mass suspended at the edge,
g be the acceleration due to gravity.

The diagram should also show:
(a) force Mg downwards at the center of the yoyo,
(b) force mg downwards at the center of the suspended mass,
(c) force T upwards in the string just above the suspended mass.

Resolving vertically for the yoyo:
T1 = Mg + T2 ...(1)

Resolving verticlally for the mass m:
T2 = mg ...(2)

Moments about the center of the yoyo:
T1 r = T2 R ...(3)

From (3):
T2 = T1 r / R ...(4)

Substituting for T2 in (1):
T1(1 - r / R) = Mg

T1 = Mg / (1 - r / R).

T1 = 0.130 * 9.81 / (1 - 1 / 4.25)
T1 = 1.67 N

From (4):
T2 = Mgr / R(1 - r / R)

T2 = Mg / (R / r - 1).

T2 = 0.130 * 9.81 / (4.25 - 1)
T2 = 0.39 N

From (2):
m = T2 / g

m = M / (R / r - 1).

m = 0.130 / (4.25 - 1)
= 0.04 kg

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