Interactive Lecture Demonstration 16 – Yo-yo

A yo-yo is shaped like a spool of thread. There is a small diameter central rod around which a string is wrapped. There are also two large diameter discs that form the case of the yo-yo. In this demonstration, we will apply Newton's Second Law to both the translational and the rotational motion of a yo-yo. The yo-yo will be released from rest at a height of about 1 meter above the floor. As the yo-yo unwinds, its center of mass will accelerate downward. At the same time, the yo-yo will also have a rotational acceleration.

The motion of the center of mass (translational motion) obeys Newton's Second Law for translation: F_net = m a_cm. The rotational motion obeys Newton's Second Law for rotation: t_net = I a.

1. Draw a force diagram for the yo-yo while it is suspended by the unwinding string. Then write the net force equation for the center of mass of the yo-yo.

2. In order to write a net torque equation for the yo-yo, we must choose a point to consider our axis of rotation. Since the yo-yo is accelerating (i.e. is not in equilibrium), we can only choose one of the following points for our axis: a) the center of mass of the yo-yo, or b) the point of contact of the yo-yo with the string. Since the yo-yo is not slipping against the unwinding string, that point of contact is always instantaneously at rest with respect to the ground, which is the reason we are allowed to use that point for our axis. We can think of the yo-yo as "rolling" without slipping down the string.

Let's choose the axis of rotation to be the center of mass of the yo-yo. Write the net torque equation for the yo-yo about that axis. Note that you must choose the positive direction for your torques to be consistent with the positive direction for your forces: a torque must be positive if it tends to rotate the yo-yo in the direction that would make the yo-yo roll down the string (the same direction as the acceleration of the yo-yo's center of mass).

3. We can write one more equation that will help us solve this problem. Since the yo-yo does not slip against the string, the tangential acceleration of the rim of the yo-yo must be equal in magnitude to the acceleration of the center of mass of the yo-yo: a_t = a_cm . Thus, using the relation between tangential and angular acceleration for any rotating object, we see that

a_cm = a r. Combine this formula with your other two equations to find a formula for the acceleration of the center of mass of the yo-yo, in terms of g, I, M and r.

4. Now examine your result. It should be clear that the yo-yo's center of mass will have less acceleration than it would in free fall, since the string's tension pulls the yo-yo upward. Does your formula show that a_cm < g ?

5. Your formula for a_cm involves the moment of inertia of the yo-yo. We are now going to estimate that quantity by using the definition of moment of inertia: I = S m_i r_i². This formula says that to calculate the moment of inertia, we have to consider the object to be made of many small pieces of mass m_i. We have to find the distance r_i of each piece of mass from the axis of rotation and then add up the quantity m_i r_i² for the whole object. Rather than actually doing that calculation (which would involve calculus), we will just use the results given in the textbook. The formula for the moment of inertia of a cylindrical object rotating around its central axis is (see p. 232 of Serway + Faughn) I = (1/2) M R². M is the mass of the object, R is its radius. The yo-yo has three parts: a thin rod around which the string is wrapped, and two wide cylindrical sides. Since the formula for moment of inertia just involves a summation over all the parts of the object, we can just add the moments of inertia of each of those three parts to get the total moment of inertia. We will consider each part to be a cylinder.

Almost all of the mass of the yo-yo is in the large sides: the central rod is very light. In addition, the radius of the central rod is much less than the radius of the sides. For both those reasons, the moment of inertia of the central rod is very small compared to that of the sides, so we are going to neglect the rod's moment of inertia. In that case, the formula for the total moment of inertia of the yo-yo becomes very simple: we can just use I = (1/2) M R², where M is the total mass of the yo-yo and R is the radius of the sides.

The measured total mass of the yo-yo is about 50 grams. The radius of the large discs is about 2.2 cm. Calculate the yo-yo's moment of inertia.

6. Now use your formula from #3 to predict the acceleration of the center of mass of the yo-yo. Note that the radius r we should use for the rod of the yo-yo should be some average radius that includes the width of the wound string. The radius of the rod + string when the yo-yo is completely wound is about 0.96 cm. The radius of the rod (with no string wound) is about 0.34 cm. The average radius we should use is thus about (0.96 cm + 0.34 cm)/2 = 0.65 cm.

7. We are going to test our formula by timing the yo-yo as it falls. Since the center of mass of the yo-yo will move with constant acceleration, we can use a DVAT. Let's say the yo-yo takes a time t to fall a distance d, starting form rest. Find a formula for the acceleration of the yo-yo.

8. Using a stopwatch, time the yo-yo as it falls from a height of 1 meter. Compare the actual time with the value predicted by your DVAT formula.