Rolling Dynamics

Acceleration

When an object rolls without slipping the angular acceleration and the linear acceleration of the object are related both in direction and magnitude. The relationship between the directions of the angular and linear accelerations is easily understood from a diagram like Fig. 1. On the left, in Fig. 1(a), a ball is rolling to the right. Naturally the ball will roll in the same direction, that is, the rotation will be clockwise. As the object speeds up, the clockwise rotation must also speed up. The rotation could never slow down as shown in Fig. 1(b). Keep this in might when drawing your free-body diagram! 

The magnitude of the linear acceleration is related to angular acceleration by a = Ra, where R is the distance from the centre of the rolling object to the surface of contact. Be careful! A rolling object can have several radii; you must choose the proper radius to use. For example consider Fig. 2 which shows a yo-yo rolling on a broad flat surface like a table or on a narrow rail that touches the inner cylinder of the yo-yo. In Fig. 2(a), the distance from the centre of the yo-yo to the table is R₁ so a₁ = R₁a₁ while in Fig. 2(b) the distance from the centre of the yo-yo to the rail is R₂ so a₂ = R₂a₂.

Friction

Friction usually plays an important role in rolling – without it objects would slide rather than roll. Since we are interested in rolling objects not sliding objects, and as the point of contact between the rolling object and the surface is instantaneously at rest, we are dealing with static friction not kinetic friction. Moreover, you definitely should not assume that you are dealing with f_s^max. There is no formula for f_s such as the formula for f_s^max = m_sN. You will need to apply Newton’s Laws to derive a formula for f_s.

Your first step in solving a rolling dynamics problem is to draw a free body diagram. Your first hurdle is to decide on the direction of f_s. There are three approaches to determining this direction:

1. Consistency with Newton’s Law’s.
2. Static friction will act to prevent slipping due to pulling and torque.
3. Arbitrarily choose a direction and check the sign of the derived formula for f_s.

In Fig. 3, we see two yo-yo’s on a flat surface. A string is tied to the central cylinder of each and this causes the yo-yo to roll to the right with increasing speed. The direction of the tension in the string, the normal force, and weight are all easy to find and are already added to the free body diagrams. The direction of the friction, either to the right or to the left, must be decided upon. Consider Fig 3(a), Newton’s Laws say  and , but we have an acceleration a₁ to the right but no force. There is a clockwise torque from the tension that can cause our clockwise acceleration a_1. This implies that f_s must point to the right for the diagram to be consistent with Newton’s Laws. 

Consistency with Newton’s Laws does not help with Fig 3(b). The tension provides both a force to the right and a clockwise torque which is consistent with a₂ and a₂. So there is no reason from Newton’s Laws at this point to say which way friction should point. The second approach, choosing friction to prevent slipping will also not work here. The tension is pulling the object, and thus the point of contact, to the right; friction would act to the left to resist this just as it would on a block. However the torque due to the tension is also trying to rotate the clockwise potentially sliding the point of contact to the left and friction would act to the right to prevent this.  We cannot say which effect is greatest, so we much simply choose a direction. When we have finished solving all the algebra, our equation for f_s will either be positive indicating we chose correctly or negative indicating that friction was opposite to our choose. There is no shame or poor physics in guessing wrong and getting a negative value for f_s just in not explaining that you know that you did choose the wrong direction.

Let’s work though an example.

A yo-yo of mass M and moment of inertia I = ³/₅MR² has outer radius R and inner radius r = ¹/₂R. A string is wrapped around the inner cylinder and pulled with tension T to the right. As a result the yo-yo rolls with increasing speed to the right. The coefficients of friction between yo-yo and the surface are m_s and m_k.

(a) Find the acceleration of the yo-yo in terms of the given variables.

(b) Find the friction acting on the yo-yo in terms of the given variables.

(c) It the string is pulled too hard, the yo-yo will begin to slip. Find an expression for the tension at which the yo-yo will first start to slip.

The free-body diagram will be much like Fig 3(b) but let’s guess that f_s acts to the right.

Applying Newton’s Laws we find:

N – Mg = 0                                       (1)

T + f_s = Ma                                       (2)

Rf_s – ¹/₂RT = –(³/₅MR²)a                 (3)

We also know

    a = a/R                                    (4)

We can use Eq. 4 in Eq. 3 to eliminate a since we are not interested in that quantity. Doing so yields

        Rf_s – ¹/₂R T = –³/₅MRa                      (3b)

We can use Eq. 2 to isolate f_s

                                                                    f_s =  Ma – T                                                                          (2b)

and substitute the result into Eq. 3b to get rid of f_s

                                                          ( Ma – T) – ¹/₂T = –³/₅M a                                                  (3b)

When we collect all the terms with a to one side and solve we get

                                                                         (5)

Which is only in terms of variables given in the question and is thus the answer to (a). In turn we can get an expression for f_s by substituting Eq. (5) into Eq. (2b) which yields

                                                                      f_s = –¹/₁₆ T                                                                         (6)

The minus sign indicates that the friction should have pointed left. So the answer to (b) is f_s = ¹/₁₆T directed left.

Note that as T increases so does f_s. For part (c), the yo-yo will slip if our equation for f_s matches f_s^max. We know that  f_s^max = m_sN = m_sMg using Eq (1). So the slipping will start when ¹/₁₆T = m_sMg or when T = 16m_sMg.

Summary

• a and a must have consistent directions
• a = Ra where R is the distance from the centre of the object to the surface.
• Use f_s – not f_k, not f_s^max.
• Use Newton’s Laws to determine the direction of f_s if possible. If not, see if one direction of f_s will resist slipping for both pulling and rotating. If all else fails, choose one direction but pay attention to the sign of the result for f_s.
• Your final equations for a and f_s must only be in terms of given quantities, i.e. the variables mentioned in the problem and g.
• An object will start to slip when f_s = f_s^max. Use your formula for f_s for the left hand side and m_sN for the right hand side.