Question

The slotted link is pinned at O, and as a result of the constant angular velocity ??= 3 rad / s it drives the peg P for a short distance along the spiral guide r= ( 0.4 ? ) m, where ? is in radians.

0.5 m θ-3 rad /s

Part A) Determine the radial component of the velocity of P at the instant ?=? / 3rad.

Part B) Determine the transverse component of the velocity and acceleration of P at the instant ?=? / 3rad.

Part C) Determine the radial component of the acceleration of P at the instant ?=? / 3rad.

Part D) Determine the transverse component of the acceleration of P at the instant ?=? / 3rad.

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Answer #1
Concepts and reason

When a particle moves along a straight line path then it is known as rectilinear motion.

When a particle moves along a curved path it is known as curvilinear motion.The time derivative of position of the particle is known as velocity. The time derivative of velocity is known as acceleration.

In cylindrical coordinates, the specification of position of the particle is in terms of radial coordinate r and a transverse coordinate. The directions r and are perpendicular to each other.

Fundamentals

The velocity of a particle in cylindrical coordinates is,

v=vu, + vyu

Here, is radial velocity component and is transverse component.

The radial component of velocity is rate of increase or decrease in length of the radial coordinate.

The transverse component is rate of motion along the circumference of the circle.

Ve=re

The magnitude of the velocity of the particle is,

201+z+/=
24+ (a1 = 1

The acceleration of a particle is,

a=au, +a,u,

The radial component of acceleration is,

201-1= o

The transverse component of acceleration is,

a = rᎾ +2rᎾ

The magnitude of the acceleration is,

a= Ja ta
= Nr-10>)* +(r0+2ro)

Consider the radial distance moved by the particle as,

r=0.40
……. (1)

Here, is the short distance along the spiral guide, is the angle turned by slotted link.

Differentiate equation (1), with respect to t.

……. (2)

Substitute 0.40 m
for .

d(0.40 m)
= 0.40 m/s

Substitute 3 rad/s
for .

r=0.40 m/s
= 0.4x3
= 1.2 m/s

Differentiate equation (2), with respect to t.

Substitute (0.40)
for .

d(0.40 m )
dt
= 0.40 m/s

Substitute O rad/s?
for .

r=0.4(0 rad/s)
= 0 m/s

A)

Calculate the radial component of velocity of the particle.

V, =r
= 1.2 m/s

B)

Calculate the transverse component of velocity of the particle.

Od = 1

Substitute3 rad/s.
for and for .

Ve=0.40x3
= 1.20 m/s

Substitute for .

vo=1.2(1)
= 1.2566 m/s

C)

Calculate the radial component of acceleration of the particle.

a, =r-r(0)

Substitute 0 for , for and for .

a, = 0–0.40x(32)
--04()*(3)
= -3.77 m/s2

D)

Calculate the transverse component of acceleration of the particle.

a = rᎾ +2rᎾ

Substitute 1.2 m/s
for , 0 for, for and for .

a, =0.40x0+2x1.2x3
= 7.2 m/s

Ans: Part A

Therefore, the radial component of velocity of the particle is 1.2 m/s
.

Part B

Therefore, the transverse component of velocity of the particle is 1.2566 m/s

Part C

Therefore, the radial component of acceleration of the particle is -3.77 m/s?
.

Part D

Therefore, the transverse component of acceleration of the particle is 7.2 m/s

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