Examples of Solutions to Problems Involving Motion of Rigid Bodies (Part - 2) Civil Engineering (CE) Notes | EduRev

General discussion of slipping contacts: 

Solving problems with sliding at a contact is always tricky, because we have to draw the friction forces in the correct direction.   Before tackling the example, we will summarize the general rules.

We will consider a wheel as an example, but the rules apply to contact between any object and a stationary surface. The figure shows a wheel that spins with angular velocity ω = ω _zk while the center moves with speed v_O = v_oxi .

The direction of the friction force is determined by the direction of motion of the point on the wheel that instantaneously touches the ground, which can be calculated from the formula

v_C = (v_Ox + ω_zR)i

Friction always acts to try to bring point C to rest – if C is moving to the right, friction acts to the left; if C is moving to the left, friction acts to the right.

There are three possible cases:

Forward slip: v_Ox + ω _zR> 0 Point C moves in the positive i direction over the ground

• Slip occurs at the contact,

• We have to use the friction law T = µN
• Point C is moving to the right, so friction must act to the left

Pure rolling v_Ox + ω _zR= 0 .  Point C is stationary. 

• No slip occurs at the contact.  
• In this case|T| < µ N
• We can draw the friction force in either direction at the contact (if we choose the wrong direction, our calculations will just tell us that T is negative).  It is usually convenient to choose T to act in the positive i direction, but this is not necessary.

Reverse slip: v_Ox + ω_zR<0 Point C moves in the negative i direction over the ground

• Slip occurs at the contact,
• We have to use the friction law T = µN
• Point C is moving to the left, so friction must act to the right

Now we return to the example.

4.1 Draw a free body diagram showing the forces acting on the disk just after it hits the ground.  

We are given that v_x0 and ω_z0 are both positive so we have v_Ox + ω_zR> 0 .  This is forward slip, and we use the corresponding FBD.

4.2 Hence, find formulas for the initial acceleration a and angular acceleration α for the disk, in terms of g , R and µ .  Note that the contact point is slipping.

The equations of motion are 

-μNi + (N - mg) j = ma_x i

- μNRk = I_Gzzα_z k

Solving these and using I_Gzz = mR² /2 :

4.3 Find formulas for the velocity and angular velocity of the disk, during the period while the contact point is still slipping.

The acceleration and angular acceleration are constant, so we can use the constant acceleration formulas:

v_x = v₀ − µgt ω_z =ω₀ − 2µgt/ R

4.4 Find a formula for the time at which the disk will reverse its direction of motion.

      Velocity is reversed where v=0.   From the previous part, v = v0 − µgt ⇒ t= v₀ /µg at the reversal.

4.5 Find a formula for the time at which the disk begins to roll on the ground without slip.   Hence, show that the disk will reverse its direction only if v₀< ω₀R /2

      Rolling without sliding starts when v_xO = −ω_zR .   We have that 

The reversal will only occur if rolling without slip occurs after the reversal of velocity. This means

(v₀ +ω₀R) /3µg >v₀ /µg ⇒ v₀<ω₀R / 2

Example 5: The ‘Sweet Spot’ on a softball or baseball bat, or tennis or squash racket is a point that minimizes the reaction forces acting on the athlete’s hand when the ball is struck.   In fact, any rigid body has a sweet spot – the magic point is called the ‘center of percussion’ of a rigid body.

For baseball and softball bats in particular, there is a standard ASTM test that can be used to measure the position of the sweet spot.   The test works like this: the bat is suspended from the knob on handle, so it swings like a pendulum.   The period of vibration τ of the swinging bat is then measured.  ASTM say that the center of percussion is then a distance

d = π²g/4π²

from the end of the handle.   Why does this work?   It seems that this test has nothing whatever to do with a ball hitting the bat!

We will solve this problem in two parts.  First, we will calculate a formula for the period of vibration in the ASTM test.   Then we will calculate the position of the center of percussion.  We will see that the ASTM test does indeed make the correct prediction.

We can calculate the period using the energy method.  The figure shows the ASTM pendulum test.  We assume that

• The bat has a mass moment of inertia about its COM I_Gzz 
• The COM is a distance L from O

The bat pivots about O, so we can use the fixed axis rotation formula for the kinetic energy

Here I_Ozz = I_Gzz + ML² (using the parallel axis theorem). 

The potential energy is V = − MgL cosθ 

Energy conservation gives

If θ is small then sinθ ≈θ so the equation of motion reduces to

This is a standard ‘Case 1’ EOM.  The natural frequency is   so the period is

Next, we find the position of the ‘sweet spot’. We can do this by calculating the reaction forces on the handle when the bat is struck, and finding the impact point that minimizes the reaction force.  

The figure shows an impact event.   We assume that:

• The bat rotates in the horizontal plane (so gravity acts out of the plane of the figure).  
• The bat rotates about the point O
• The ball impacts the bat a distance d from the handle.
• The ball exerts a (large) force F_impact on the bat
• Reaction forces R_x ,R_y act on the handle during the impact.

This is a planar problem, so we can use the 2D equations of motion. The equation for translational motion gives

( R_x −F_impact)i + R_y j= Ma_G 

For the rotational equation we can also use the short-cut for fixed axis rotation

We can relate a_G to α_z using the rigid body formula:

a_G =α _zk ×r_G−ω² _z rG =α_z Li +ω² _zLj

We therefore see that

The sweet spot is at the position that makes R_x = 0 , which shows that

For comparison, the ASTM formula gives