Unit - 4

Review of particle dynamics

Q1) The car in figure moves in a straight line such that for a short time its velocity is defined by ft/s where t is in seconds. Determine its position and acceleration when and if , .

A1)

Coordinate system

The position coordinate extends from the fixed origin O to the car, positive to the right.

Position.

Since This equation relates Nothing that when we have*

When

Acceleration.

Since the acceleration is determined from .

Since this equation relates

When t=3s,

Q2) At any instant the horizontal position of the weather balloon is defined by where t is in seconds and x is in meters. If the equation of the path of the balloon is . Determine the magnitude and direction of the velocity and the acceleration when t = 2 s.

A2) Velocity:

The velocity component in the x direction is

At t = 2 s

The velocity component in the y direction is

At t = 2 s

The direction is tangent to the path

Acceleration:

Magnitude of acceleration is

Direction of acceleration is

Q3) For a short time, the path of the plane is described by If the plane is rising with a constant velocity of 10 m/s, determine the magnitudes of the velocity and acceleration of the plane when it is at y = 100 m.

A3) When or Also, since , then

Velocity

Relationship between the velocity components, we have

Thus

The magnitude of the velocity is therefore

Acceleration: Using the chain rule, the time derivative of Eq. (1) gives the relation between the acceleration components.

When

The magnitude of the plane’s acceleration is therefore

Q4) A race car C travels around the horizontal circular track that has a radius of 300 ft. If the car increases its speed at a constant rate of 7 ft/s2 starting from rest, determine the time needed for it to reach an acceleration of 8 ft/s2. What is its speed at this instant?

A4) Coordinate System

The origin of the n and t axes is coincident with the car at the instant considered. The t axis is in the direction of motion, and the positive n axis is directed toward the center of the circle. This coordinate system is selected since the path is known.

Acceleration

The magnitude of acceleration can be related to its components using . Here . Since . The velocity as a function of time must be determined first.

Thus

The time needed for the acceleration to reach is therefore

Solving for the positive value of t yields

Velocity. The speed at time is

Q5) The amusement park ride shown in Fig below consists of a chair that is rotating in a horizontal circular path of radius r such that the arm OB has an angular velocity and angular acceleration . Determine the radial and transverse components of velocity and acceleration of the passenger. Neglect his size in the calculation.

A5)

Coordinate system

Since the angular motion of the arm is reported, polar coordinates are chosen for the solution. Here θ is not related to r, since the radius is constant for all θ.

Velocity and Acceleration:

It is first necessary to specify the first and second time derivatives of r and θ. Since r is constant, we have

Thus

Q6) The 50-kg crate shown in Fig below rests on a horizontal surface for which the coefficient of kinetic friction is If the crate is subjected to a 400-N towing force as shown, determine the velocity of the crate in 3 s starting from rest.

A6)

Using the equation of motion, we can relate the crate’s acceleration to the force causing the motion. The crate’s velocity can then be determined using kinematics.

Free-Body Diagram:

The weight of the crate is

As shown in Fig. the frictional force has a magnitude and acts to the left, since it opposes the motion of the crate. The acceleration a is assumed to act horizontally, in the positive x direction. There are two unknowns, namely and a.

Equations of motion

      (1)

     (2)

Solving Eq.2 for , substituting the result into Eq.1 and solving for  a yields

Kinematics

Notice that the acceleration is constant, since the applied force P is constant. Since the initial velocity is zero, the velocity of the crate in 3 s is

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Q7) A smooth 2-kg collar C, shown in Fig, is attached to a spring having a stiffness k = 3 N/m and an unstretched length of 0.75 m. If the collar is released from rest at A, determine its acceleration and the normal force of the rod on the collar at the instant y = 1 m

A7) Free-Body Diagram:

The free-body diagram of the collar when it is located at the arbitrary position y is shown in Fig. Furthermore, the collar is assumed to be accelerating so that “a” act downward in the positive y direction. There are four unknowns, namely and θ.

Equation of Motion:

        (1)

+       (2)

From Eq. 2 it is seen that the acceleration depends on the magnitude and direction of the spring force. Solution for and a is possible once  and θ are known.

The magnitude of the spring force is a function of the stretch s of the spring; i.e.,         . Here the unstretched length is Fig. 13-19a; therefore,

. Since , then

    (3)

From Fig 13-99a, the angle θ is related to y by trigonometry.

         (4)

Substituting into Eqs. 3 and 4 yields and . Substituting these results into Eqs. 1 and 2, we obtain

Q8) Determine the banking angle for the race track so that the wheels of the racing cars will not have to depend upon friction to prevent any car from sliding up or down the track. Assume the cars have negligible size, a mass m, and travel around the curve of radius with a constant speed v.

A8) Free-Body Diagram:

As shown in Fig. and as stated in the problem, no frictional fore acts on the car. Here represents the resultant of the ground on all four wheels. Since can be calculate, the unknowns are and θ.

Equations of Motion;

Using the n, b axes shown,

      (1)

+       (2)

Eliminating and m from these equations by dividing Eq.1 by Eq.2 we obtain

Q9) The 3-kg disk D is attached to the end of a cord as shown in Fig. The other end of the cord is attached to a ball-and-socket joint located at the center of a platform. If the platform rotates rapidly, and the disk is placed on it and released from rest as shown, determine the time it takes for the disk to reach a speed great enough to break the cord. The maximum tension the cord can sustain is 100 N, and the coefficient of kinetic friction between the disk and the platform is

A9)

Free-Body Diagram:

The frictional force has a magnitude and a sense of direction that opposes the relative motion of the disk with respect to the platform. It is this force that gives the disk a tangential component of acceleration causing v to increase, thereby causing T to increase until it reaches 100n. the weight of the disk is . Since can be related to v, the unknowns are and v.

Equations of Motion:

         (1)

         (2)

        (3)

Setting , Eq. 1 can be solved for the critical speed of the disk needed to break the core. Solving all the equations, we obtain

Kinematics:

 Since is constant, the time needed to break the cord is

Q10) The smooth 0.5-kg double-collar can freely slide on arm AB and the circular guide rod. If the arm rotates with a constant angular velocity of , determine the force the arm exerts on the collar at the instant 450. Motion is in the horizontal plane.

A10) Free-body diagram: The normal reaction of the circular guide rod and the force F of arm AB act on the collar in the plane of motion. Note that F acts perpendicular to the axis of arm AB, that is, in the direction of the θ axis, while acts perpendicular to the tangent of the circular path at . The four unknowns are .

Equations of Motion:

  (1)

  (2)

Kinematics:

Using the chain rule (see Appendix C), the first and second time derivative of  r when are

]

We have

Substituting these results into Eqs. (1) and (2) and solving, we get

Q11) The man in Figure pushes on the 50-kg crate with a force of Determine the power supplied by the man when . The coefficient of kinetic friction between the floor and the crate is . Initially the create is at rest.

A11)

To determine the power developed by the man, the velocity of the 150-N force must be obtained first. The free-body diagram of the crate is shown in Fig. applying the equation of motion

The velocity of the crate when is therefore

The power supplied to the crate by the man

When t=4s is therefore

Q12) For a short time the crane lifts the 2.50-Mg beam with a force of Determine the speed of the beam when it has risen. Also, how much time does it take to attain this height starting from rest?

A12) We can solve part of this problem using the principle of work and energy since it involves force, velocity, and displacement. Kinematics must be used to determine the time. Note that at , so motion will occur.

Work (Free-Body Diagram). As shown on the free-body diagram, the lifting F does positive work, which must be determined by integration since this force is a variable. Also, the weight is constant and will do negative work since the displacement is upwards.

Principle of work and Energy.

When

Kinematics. Since we were able to express the velocity as a function of displacement, the time can be determined using = ds/dt. In this case,

The integration can be performed numerically using a pocket calculator. The result is

Q13) Blocks A and B shown in Figure below have a mass of 3 kg and 5 kg, respectively. If the system is released from rest, determine the velocity of block B in 6 s. Neglect the mass of the pulleys and cord.

A13) Free-Body Diagram. Since the weight of each block is constant, the cord tensions will also be constant. Furthermore, since the mass of pulley D is neglected, the cord tension . Note that the blocks are both assumed to be moving downward in the positive coordinate directions, and .

Principle of Impulse and Momentum.

Block A:

(+)   

Block B:

(+)   

Kinematics. Since the blocks are subjected to dependent motion, the velocity of A can be related to that of B by using the kinematic analysis a horizontal datum is established through the fixed point at C, Fig. 15-6a, and the position coordinates, and , are related to the constant total length of the vertical segments of the cord by the equation

Taking the time derivative yields

    (3)

As indicated by the negative sign, when B moves downward A moves upward. Substituting this result into Eq. 1 and solving Eq. 1 and 2 yields

Q14) Define Impact and state its types.

A14) Impact occurs when two bodies collide with each other in a very short period of time, resulting relatively huge (impulsive) forces to be exerted between the bodies.

Example: striking of a hammer on a nail, a golf club on a ball

There are two types on impact:

Central impact occurs when the direction of motion of the mass centers of the two colliding particles is along a line passing through the mass centers of the particles. This line is called the line of impact which is perpendicular to plane of contact.

2.     Oblique impact

When the motion of one or both of the particles make an angle with the line of impact, then the impact is said to be oblique impact.