Chapter 17. Planar Kinetics of a Rigid Body

Transcription

1 Chapter 17 Planar Kinetics of a Rigid Body Force and Acceleration Engineering Mechanics : Dynamics R.C. Hibbeler

2 MOMENT OF INERTIA The mass moment of inertia is a measure of an object s resistance to rotation. Thus, the object s mass and how it is distributed both affect the mass moment of inertia. Mathematically, it is the integral I = r 2 dm = r 2 ρ dv In this integral, r acts as the moment arm of the mass element and r is the density of the body. Thus, the value of I differs for each axis about which it is computed. The figures below show the mass moment of inertia formulations for two flat plate shapes commonly used when working with three dimensional bodies. The shapes are often used as the differential element being integrated over the entire body. PARALLEL-AXIS THEOREM If the mass moment of inertia of a body about an axis passing through the body s mass center is known, then the moment of inertia about any other parallel axis may be determined by using the parallel axis theorem, I = I G + md 2 where IG = mass moment of inertia about the body s mass center m = mass of the body d = perpendicular distance between the parallel axes Radius of Gyration The mass moment of inertia of a body about a specific axis can be defined using the radius of gyration (k). The radius of gyration has units of length and is a measure of the distribution of the body s mass about the axis at which the moment of inertia is defined. I = mk 2 or k = I / m

3 Kinetics of a Rigid Body: Force and Acceleration Equation of translational motion We will limit our study of planar kinetics to rigid bodies that are symmetric with respect to a fixed reference plane. As discussed in Chapter 16, when a body is subjected to general plane motion, it undergoes a combination of translation and rotation. First, a coordinate system with its origin at an arbitrary point P is established. The x-y axes should not rotate and can either be fixed or translate with constant velocity. If a body undergoes translational motion, the equation of motion is ΣF = ma G. This can also be written in scalar form as Σ F x = m(a G ) x and Σ F y = m(a G ) y In words: the sum of all the external forces acting on the body is equal to the body s mass times the acceleration of it s mass center. = Equation of rotational motion We need to determine the effects caused by the moments of the external force system. The moment about point P can be written as Σ (r i F i ) + Σ M i = r G ma G + I G α Σ M p = Σ( M k ) p where Σ Mp is the resultant moment about P due to all the external forces. The term Σ(M k ) p is called the kinetic moment about point P. If point P coincides with the mass center G, this equation reduces to the scalar equation of Σ M G = I G α. In words: the resultant (summation) moment about the mass center due to all the external forces is equal to the moment of inertia about G times the angular acceleration of the body.

4 Thus, three independent scalar equations of motion may be used to describe the general planar motion of a rigid body. These equations are: Σ F x = m(a G ) x Σ F y = m(a G ) y and Σ M G = I G α or Σ M p = Σ (M k ) p Equation of Motion : Translation only When a rigid body undergoes only translation, all the particles of the body have the same acceleration so a G = a and α = 0. The equations of motion become: Σ F x = m(a G ) x Σ F y = m(a G ) y Σ M G = 0 Note that, if it makes the problem easier, the moment equation can be applied about other points instead of the mass center. In this case, ΣM A = (ma G ) d. When a rigid body is subjected to curvilinear translation, it is best to use an n-t coordinate system. Then apply the equations of motion, as written below, for n-t coordinates. Σ F n = m(a G ) n Σ F t = m(a G ) t Σ M G = 0 or Σ M B = e[m(a G ) t ] h[m(a G ) n ]

5 Equation of motion for pure rotation When a rigid body rotates about a fixed axis perpendicular to the plane of the body at point O, the body s center of gravity G moves in a circular path of radius r G. Thus, the acceleration of point G can be represented by a tangential component (a G ) t = r G a and a normal component (a G ) n = r G ω 2. Since the body experiences an angular acceleration, its inertia creates a moment of magnitude I G α equal to the moment of the external forces about point G. Thus, the scalar equations of motion can be stated as: F n = m (a G ) n = m r G ω 2 F t = m (a G ) t = m r G α M G = I G α Note that the M G moment equation may be replaced by a moment summation about any arbitrary point. Summing the moment about the center of rotation O yields M O = I G α + r G m (a G ) t = (I G + m (r G ) 2 ) α From the parallel axis theorem, I O = I G + m(r G ) 2, therefore the term in parentheses represents I O. Consequently, we can write the three equations of motion for the body as: F n = m (a G ) n = m r G ω 2 F t = m (a G ) t = m r G α M O = I O α Equation of motion : General plane motion When a rigid body is subjected to external forces and couple-moments, it can undergo both translational motion as well as rotational motion. This combination is called general plane motion. Using an x-y inertial coordinate system, the equations of motions about the center of mass, G, may be written as F x = m(a G ) x F y = m(a G ) y M G = I G α P

6 Sometimes, it may be convenient to write the moment equation about some point P other than G. Then the equations of motion are written as follows. F x = m (a G ) x F y = m (a G ) y M P = (M k ) P In this case, (M k ) P represents the sum of the moments of I G α and ma G about point P.

7 Example A 50 kg crate rests on a horizontal surface for which the kinetic friction coefficient μk = 0.2. Find:The acceleration of the crate if P = 600 N. The handcart has a mass of 200 kg and center of mass at G. Determine the normal reaction at each of the two wheels at A and the two wheels at B if a force of P =50 N is applied to the handle. Neglect the mass of the wheels. The crate C has a weight of 1500 N and rests on the truck elevator for which the coefficient of static friction is μ s = 0.4. Determine the largest initial angular acceleration α, starting from rest, Which the parallel link AB and DE can have without causing the crate to slip. No tipping occurs. The 80-kg disk is supported by a pin at A. If it is released From rest from the position shown, determine the initial horizontal and vertical component of reaction at the pin. The 10-kg wheel has a radius of gyration k A =200 mm. If the wheel is subjected to a moment M = 5t Nm, where t is in second, determine its angular velocity where t=3 s starting from rest. Also, compute the reaction which the fixed pin A exerts on the wheel during the motion. The 20-kg roll of paper has a radius of gyration k A = 90 mm About an axis passing through point A. It is pin-support at both ends by two brackets AB. If the roll rests against a wall for which the coefficient of kinetic friction is μ k = 0.2 and a vertical force F = 30 N is applied to the end of the paper, determine the angular acceleration of the roll as the paper unrolls.

8 The uniform 500-N board is suspended from cords at C and D. If these cords are subjected to constant forces of 300 N and 450 N, respectively, determine the acceleration of the board s center and the board s angular acceleration. Assume the board is a thin plate. Neglect the mass of the pulleys at E and F. The spool has a mass of 100 kg and a radius of gyration of k g = 0.3 m. If the coefficients of static and kinetic friction at A are μ s = 0.2 and μ k = 0.15, respectively, determine the angular acceleration of the spool if P =50 N The uniform slender pole has a mass of 100 kg and a moment of inertia I G = 75 kg m 2. If the coefficients of static and kinetic friction between the end of the pole and the surface are μ s = 0.3 and μ k = 0.25, respectively, determine the pole s angular acceleration at the instant the 400 N horizontal force is applied. The pole is originally at rest. [see example 17.16]