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The boy of mass 40 kg is sliding down the spiral slide at a constant speed...

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The boy of mass 40 kg is sliding down the spiral slide at a constant speed such that his position, measured from the top of the chute, has components r = 1.5 m, θ = (0.7t) rad, and z = (-0.5t) m, where t is in seconds.
Determine the components of force Fr, Fθ, and Fz
which the slide exerts on him at the instant t = 2 s.
Neglect the size of the boy.

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Concepts and reason

Equation of motion:

It refers to a set of equations that relates the kinematic variables like displacement, velocity and acceleration of a moving object as a function of time.

Newton’s second law:

It states that “the force applied to a body in a direction is directly proportional to the mass and acceleration of the body along that direction”.

Angular displacement:

Angular displacement is the angle through which an object moves on a circular path. It is denoted by .

Velocity:

Velocity is the change in displacement with respect to time. It is denoted by and its unit is .

Acceleration:

Acceleration of an object is the change in velocity with respect to time. It is denoted by and its unit is .

Angular velocity:

Angular velocity is the change in the angular displacement with respect to time. It is denoted by or . Its unit is .

Angular acceleration:

Angular acceleration is the change in angular velocity with respect to time. It is denoted by or . Its unit is rad/s2
.

Acceleration can be calculated by differentiating displacement twice with respect to time. Then, force in each direction can be computed by applying equilibrium force equations in respective direction from Newton’s second law of motion.

Fundamentals

The angular velocity is expressed as follows:

Here, change in angular displacement is and the change in time is .

Polar coordinates:

A ball travelling on a curved path with its location in terms of polar coordinates is shown as in Figure (1).

Path
Figure 1

Here, angle made by the ball with the x-axis is , radial distance is r, unit vector established along the positive r-axis is and unit vector established along the positive -axis is .

The location of the ball in vector form at point A is written as,

r=re,

Write the velocity of the ball in polar coordinates.

v=re, +rde,

Here, first derivative of radial distance with respect to time is , first derivative of unit vector is .

Write the acceleration of the ball in polar coordinates.

a = a, e, +a,e,

a=(-ro?), +(rö +2r0 )e,
…… (1)

Here, the acceleration of ball in radial direction is , the acceleration of ball in direction is and second derivative of radial distance with respect to time is .

In a spiral system, a vertical displacement in z direction of object occurs in addition to polar coordinates system. Then, linear acceleration in z direction is calculated as follows:

Here, displacement in z direction is .

Equations of motion:

When an object with mass is subjected to concurrent forces action whose vector summation is , then equation of motion can be written as

ΣF = ma

Here, the acceleration by the mass is .

If the motion of a ball is expressed in spiral system, then equations of motion are as follows:

Write the equilibrium force equation in r direction.

ΣΕ = ma,

Here, the total force acting in r direction is .

Write the equilibrium force equation in direction.

ΣF = ma,

Here, the total force acting in direction is .

Write the equilibrium force equation linear z direction.

ΣΕ = ma.

Here, the total force acting in z direction is .

Calculate the first derivative of radial distance.

Here, radial distance of spiral is .

Substitute for .

1-d(1.5 m)
r=0

Calculate the second derivative of radial distance.

Substitute for .

řad (1.5 m)
P=0

Calculate the first derivative of angular displacement.

Here, angular position is .

Substitute for.

o-d(0.71)
1 dt
0=0.7 rad/s

Calculate the second derivative of angular displacement.

ödo

Substitute for.

0=0
(220),p.

Calculate the first derivative of -component.

N
.

Here, linear displacement in z direction is .

Substitute for.

d(-0.51)
dt
ż=-0.5 m/s

Calculate the second derivative of -component.

2 = (-0.5)
2 = 0

Calculate the acceleration of the boy in radial direction.

a, =P-r02

Substitute for , for , and 0.7 rad/s
for .

a,= (0)-[1.5 mx(0.7 m/s)]
a, = -0.735 m/s?

Calculate the acceleration of the boy in transverse direction.

ag = rö+2rė

Substitute for , for , for , and 0.7 rad/s
for .

ag = (1.5mx0)+2(0x0.7 rad/s)
ag=0

Calculate the acceleration of the boy in-direction.

Substitute for .

a
= 0

Write the equilibrium force equation in r direction.

ΣΕ = ma,

F, = ma,

Here, the force in the radial direction is .

Substitute for and -0.735 m/s2
for.

F, = 40 kg x(-0.735 m/s²)x-
IN
1 kg.m/s
F, = -29.4 N

Negative sign indicates that the force draws the body inwards.

Calculate the component of force in transverse direction.

F = ma,

Here, the force in the transverse direction is .

Substitute for and 0 for.

Fo = (40 kg x0)
F=0N

Draw the free body diagram of boy as shown in Figure (2).

mg
F.
Figure 2

Here, mass of the boy is , acceleration due to gravity is and the force experienced by boy in z direction is .

Write the equilibrium force equation in z direction from the Figure (2) using Newton’s second law.

ΣF = ma.
F. - mg = ma.

Substitute for , 9.81 m/s
for , and 0 for .

F.-(40 kgx9.81 m/s?)*
IN
1 kg-m/s2
/2 = (40 kg x0)
F = 392.4 N

Ans:

The component of force in radial directionis -29.4 N
.

The component of force in transverse directionis .

The component of force in -direction is 392.4 N
.

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