2
KINETICS OF PARTICLES & RIDGID BODIES
Three 20 kg packages rest on a belt which passes over a pulley and is attached to a 40 kg block. Knowing that the coefficient of friction between the belt and the horizontal surface and also between the belt and the packages is 0.50, determine the speed of package B as it falls off the belt at E.
Assumption: Packages do not slip on belt.
Free-body diagram for a typical package.
Packages do not move ^ belt ® SFy = 0: 196.2 – Np = 0 ® Np = 196.2 N
Friction: F = mNp = 0.5(196.2) = 98.1 N
Principle of Work and Energy: U1®2 = T2 – T1
l Belt not moving initially ® T1 = 0
l T2 =
l U1®2 =
® ®
l Check on assumption of no slipping:
For 1 package: U1®2 = F(2), T1 = 0,
® 2 F = 39.24 ® F = 19.62 N
But Fmax = 98.1 > F ® Packages do not slip.
As the bracket ABC is slowly rotated, the 6 kg block starts to slide toward the spring when q=15. The maximum deflection of the spring is observed to be 50 mm. Determine the values of the coefficients of static and kinetic friction.
Free-body diagram
Static Case
SFx = 0: 58.86 sin 15 – F = 0 (i)
SFy = 0: Np – 58.86 cos 15 = 0 (ii)
Friction: F = msNp = ms(58.86 cos 15)
From (i) 58.86 sin 15 = ms (58.86 cos 15)
®
Dynamic Case
Principle of Work and Energy: U1®2 = T2 - T1
T1 = 0 (block starts from rest), T2 = 0 (at maximum deflection of spring, block is again at rest)
; Fmin = k ´1 = 0 (since spring is undeformed initially i.e. x1 = 0)
® 58.86(0.3)(sin 15 - mk cos 15) – 2 = 0
1. A 1500 kg automobile travels 200 m while being accelerated at a uniform rate from 50 to 75km/h. During the entire motion, the automobile is traveling on a horizontal road, and the rolling resistance is equal to 2 percent of the weight of the automobile. Determine (a) the maximum power required, (b) the power required to maintain a constant speed of 75 km/h.
2. A slender rod AB of negligible mass is attached to blocks A and B, each of mass m. The constant of the spring is k and the spring is undeformed when AB is horizontal. Determine the potential energy of the system with respect to (a) the spring, (b) gravity. (Place datum atB.)
3. The spring AB is of constant 1.2 kN/m and is attached to the 2 kg collar A which moves freely along the horizontal rod. The unstretched length of the spring is 250 mm. If the collar is released from rest in the position shown, determine the maximum velocity attained by the collar.
The 5 kg bar shown in the figure has a center of mass at G. If it is given an initial clockwise angular velocity of w1 = 10 rad/s when q = 90, compute the spring constant k so that it stops when q = 0. What are the horizontal and vertical components of reaction at the pin A when q=0? The spring deforms 0.1 m when q = 0.
· Forces which do work are the weight and the spring force
Þ Principle of work and energy can be used to determine the spring constant k
· Kinetic energies:
(bar stops when q = 0)
l Work done
l Principle of Work and Energy:
30 + 14.715 – 0.005 k = 0 ®
Note: T1 can be calculated alternatively as T1=
T1 =
l Ay and Ax do no work ® must be obtained from the equations of motion
Free-body diagram Kinetic diagram
See kinematics below
Fs = kx = 8943(0.1) = 894.3 N
From ® 49.05(0.3) – 894.3(0.6) = 0.6a ® a = -869.8 rad/s2
From ® Ay + 894.3 – 49.05 = -(0.3)(5)(-869.8) ®
l Kinematics: (A is fixed)
®
The 100 kg wheel shown in the figure has a radius of gyration of kG = 0.25 m about its center of mass G. If it is subjected to a clockwise couple of 20 N m as it rolls on its inner hub without slipping, determine the wheel’s angular velocity after the 20 kg block is released from rest and has fallen 0.4 m. The spring has a stiffness of k = 60 N/m and is initially unstretched when the block is released.
l Angular velocity is required ® Principle of work and energy is the most “economical” analytic tool to use.
l It is easier to analyze the system as a whole than to consider the components separately
Position
Position
l Kinetic Energies:
Work done
l Nw and R do not work since they do not move along their lines of action
l Fr does no work since wheel does not slip as it rolls
l Wheel does not slip ® wheel moves through q as block moves a distance s such that
l spring stretches ´ = rD/Ic q = (0.4)(0.571) = 0.228 m
l Principle of Work and Energy:
®
The uniform rods AB and BC are of mass 3kg and 8 kg, respectively, and collar C has mass of 4 kg. If the system is released from rest in the position shown, determine the velocity of point B after rod AB has rotated through 90o.
· Only gravity forces, which are conservative, are present
Þ Energy is conservative
MAB = 3 Kg, MBC = 8 Kg, MC = 4 Kg
· Kinematics: Rod AB:
Rod BC: Instantaneous Centre is at C Þ VB = 0.39
· Kinetic Energies
T1 =0; T2 =
· Potential Energies
V1= WBC (-0.180) + WC (-0.360) + WAB (0) = 9.81
V1=-28.253 J
V2 =WAB (-0.075) + WBC
= 9.81 [3(-0.075) + 8 (-0.345) + 4(-0.54)]
V2 = -50.472 J
· Conservation of Energy: T1 + V1 = T2 + V2
0 +(-28.253) = + (-50.472)
The 10-kg rod AB shown is confine so that its ends move in the horizontal and vertical slots. The spring has a stiffness of k = 800 N/m and is unstretched when q = 0o. Determine the speed of the slider block at B when q = 0o, if AB is released from rest when q = 30o. Neglect the mass of the slider blocks.
Position1 Position 2
· Only conservative forces (gravity and spring force) are present Þ Energy is conserved
· Rod is released from rest in Position 1
Þ (VG)1 = W1 = 0 Þ T = 0
V1= Vg1 + Ve1 = 98.1Y1 + ½ k x12
V1 = -98.1 (0.2sin30°) + ½ (800) (0.4sin30°)2
=-9,81 + 16 = 6.19 J
T2= ½
T2=
Þ T2= 5(0.2)2
V2 =
Conservation of Energy: T1 + V1 =T1 + V2