Kinematics of Rigid Bodies-exercise 6 PDF

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1 Kinematics of Rigid Bodies- exercise 6 Translation- This type of motion occurs when a line in the body remains parallel to its original orientation throughout the motion. When the paths of motion for any two points on the body are parallel lines, the motion is called rectilinear translation fig. (...

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Kinematics of Rigid Bodies- exercise 6 Translation- This type of motion occurs when a line in the body remains parallel to its original orientation throughout the motion. When the paths of motion for any two points on the body are parallel lines, the motion is called rectilinear translation fig. (1a). If the paths of motion are along curved lines, the motion is called curvilinear translation, fig. (1b).

Fig (1a)

Fig (1b)

When a rigid body is in translation, all the points of the body have the same velocity and the same acceleration at any given instant fig (2b and 2c)

vB

vA

aA

Fig 2

aB

2

In the case of curvilinear translation, the velocity and acceleration change in direction as well as in magnitude at every instant. In the case of rectilinear translation, all particles of the body move along parallel straight lines, and their velocity and acceleration keep the same direction during the entire motion.

Rotation about a fixed axis - When a rigid body rotates about a fixed axis, all the particles of the body, except those which lie on the axis of rotation, move along circular paths fig.3

Fig.3

Angular Velocity. The time rate of change in the angular position is called the angular velocity ω (omega). Since dθ occurs during an instant of time dt then

ω

dθ dt

This vector has a magnitude which is often measured in rad/s. It is expressed here in scalar form since its direction is also along the axis of rotation. Fig.4. We can refer to the sense of rotation as clockwise or counterclockwise. Here we have arbitrarily chosen counterclockwise rotations as positive.

Fig.4

3

Angular Acceleration. The angular acceleration α (alpha) measures the time rate of change of the angular velocity. The magnitude of this vector is

α

dω dt

d 2θ dt 2

The line of action of α is the same as that for ω (Fig. 4) however, its sense of direction depends on whether ω is increasing or decreasing. If ω is decreasing, then α is called an angular deceleration and therefore has a sense of direction which is opposite to ω . Rotation of a Representative Slab The rotation of a rigid body about a fixed axis can be defined by the motion of a representative slab in a reference plane perpendicular to the axis of rotation. Let us choose the xy plane as the reference plane and assume that it coincides with the plane of the figure, with the z axis pointing out of the paper

Velocity of any point P of the slab

v

rω

Acceleration of any point P of the slab

a

at t

an n

at

rα

an

rω 2

4

The tangential component at

points in the counterclockwise direction if the

scalar α is positive, and in the clockwise direction if negative. The normal component an always points in the direction opposite to that of r , that is, toward O .

Sample problems: 1. A cord is wrapped around a wheel in fig. which is initially at rest when θ 0 . If a force is applied to the cord and gives it an acceleration a (4t ) m / s 2 , where t is in seconds, determine, as a function of time: a) The angular velocity of the wheel; b) The angular position of line OP in radians.

Solution: a) The wheel is subjected to rotation about a fixed axis passing through point O . Thus, point P on the wheel has motion about a circular path, and the acceleration of this point has both tangential and normal components. The 4t m / s 2 , since the cord is wrapped around tangential component is aP t the wheel and moves tangent to it. Hence the angular acceleration of the wheel is:

aP α

t

α .r ;

4t m / s 2

α . 0, 2 m;

20t s 2 .

Using this result, the wheel’s angular velocity ω can now be determined from dω , since this equation relates α ,t and ω . Integrating, with the initial α dt condition that t0 0; ω0 0 , yields:

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dω dt

α

dω From initial conditions

20t s 2 ; ω

20t dt ;

C

dω

20t s 2 .dt ;

t2 20 2

10t 2 s

ω 10t 2 s

0;

1

C;

1

b) Using this result, the angular position θ of OP can be found from ω since this equation relates θ , ω and t . Integrating, we have:

dθ dt dθ

From initial conditions: t0

ω 10t 2 ; 2

10t dt ;

0;θ0

0; θ

dθ θ

C

10t 2 .dt ; t3 10 3

C.

0

t3 10 rad . 3

2. The motor shown in the picture is used to turn a wheel. If the pulley A connected to the motor begins to rotate from rest with a constant angular acceleration of α A 2 s 2 , determine the magnitudes of the velocity and acceleration of point P on the wheel, after the pulley has turned two revolutions. Assume the transmission belt does not slip on the pulley and wheel.

dθ , dt

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Solution: Angular Motion: First we will convert the two revolutions to radians. Since there are 2π rad in one revolution, then

θA

2.2π

4π

4.3,14 12,56 rad

Since α A is constant, the angular velocity of pulley A is therefore

αA

2 s 2; α A dω ω dθ

dω ; dt

dω dt

dω

2.dt ; 2t C1 ; ω

2 s 2;

2 dt ;

dθ ; dt

dθ dt dθ

2t C1 dt ;

2t C1 ; 2t dt

C1 dt ;

t2 θ 2 C1t C2 2 C1 ; C2 Constant of Integration Initial Conditions: t0

C1

C2

0; θ0

0; ω0

0

0

θ

t 2 ; When θ

ω

2.t

12,56 rad

2.3,54 7, 09 s

t

12,56

3,54s;

1

The belt has the same speed and tangential component of acceleration as it passes over the pulley and wheel. Thus,

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v ω A rA ωB αt

ωB rB ;

7, 09 s 1. 0,15 m 0, 4 m

α A .rA αB

ωB . 0, 4 m ;

7, 09 s 1. 0,15 m 2, 659 s 1 ;

α B .rB ;

2 s 2 . 0,15 m

2 s 2 . 0,15 m 0, 4 m

0, 75 s 2 .

α B . 0, 4 m

Motion of P : As shown on the kinematic diagram we have:

vP αP αP

ωB rB

2, 659 s 1. 0, 4 m

t

α B rB

n

ωB2 rB

Thus α P

0, 75 s 2 . 0, 4 m 2, 659 s αP

2 t

αP

2 2

2 n

1, 06 m / s; 0,3 m / s 2 ;

. 0, 4 m

2,827 m / s 2 ;

0,3 m / s 2

2

2,827 m / s 2

2

2,84 m / s 2 .

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3. For a short period of time, the motor turns with a constant angular gear A acceleration of α A 4, 5 s 2 , starting from rest. Determine the velocity of the cylinder and distance it travels in three seconds. The cord is wrapped around pulley D which is rigidly attached to gear B.

Solution:

Velocity vC :

vP αA

ω A . 75 mm

ωB . 225 mm ;

dω A ; dω A dt ω A α A .t C1 ;

C1

α A .dt ;

dω A

α A .dt ;

Constant of integration;

From Initial Condition :t0 ωA

α A .t ; When t

(13,5 s ). 75 mm ωB . 125 mm

vC

vP '

ωA

0

(4,5 s 1 ). 125 mm

0,5625 m / s

C1

0;

(4,5 s 2 ).3 s 13,5 s 1.

ωB . 225 mm ;

1

vP '

3s

0; ωo

ωB

(13, 5 s 1 ). 75 mm 225 mm

562,5 mm / s

4,5 s 1 ;

0,5625 m / s;

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Distance sC :

From vC

vP '

ωB . 125 mm

From ω A . 75 mm From ω A dθ A

θ B . 125 mm ;

sC

ωB . 225 mm dθ A dt

α A .t and ω A α A t.dt ; dθ A

θ A . 75 mm dθ A dt

α A t.dt ;

θ B . 225 mm ;

α A t; t2 αA 2

θA

C3 ;

Constant of integration;

C3

From Initial Condition :t0 θA

t2 α A ; When t 2

0; θ o

3s

θB

20, 25 rad . 75 mm 225 mm

sC

6, 75 rad . 125 mm

θA

0

C3 4,5 s

2

0; 32 . 2

6, 75 rad ; 843, 75 mm 0,844 m;

PROBLEMS:

1. The bucket is hoisted by the rope that wraps around a drum wheel. If the angular displacement of the wheel is θ

0,5 t 3 15 t rad , where t is in seconds,

determine the velocity and acceleration of the bucket when t 3 s.

20, 25 rad ;

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2. A motor gives gear A an angular acceleration of α A in seconds. If this gear is initially turning at ω A0 velocity of gear B when t

4t 3 rad / s 2 , where t is

20 s 1 , determine the angular

2s.

3. Starting from rest when s 0 , pulley A is given a constant angular acceleration aC 6 s 2 . Determine the speed of block B when it has risen s

6 m . The pulley has an inner hub D which is fixed to C and turns with it....