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The 0.2 kg ball is blown through the smooth vertical circular tube whose shape is defined...

The 0.2 kg ball is blown through the smooth verticThe 0.2 kg ball is blown through the smooth vertical circular tube whose shape is defined by r=(0.6sin?) m, where ? is in radians. If ?=(?t2) rad, where t is in seconds, determine the magnitude of force F exerted by the blower on the ball when t = 0.5 s.

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Concepts and reason

According to Newton’s second law of motion the rate of change of momentum of the body must be equal to the net force applied to the body in the direction of motion. In simple words, the net force acting on a body must be equal to the mass of the body multiplied by the acceleration of the body.

Similarly, Newton’s second law for rotation states that the net torque acting on the body is equal to the mass moment of inertia of the body multiplied by the angular acceleration of the body.

When all the supports are removed by replacing them with forces that prevents the translation of body in a given direction that diagram is called free body diagram. When their resultant force and couples becomes equal to zero then the body is said to be in equilibrium.

To apply these equilibrium equations, we need to know the known and unknown forces that act on the body.

In this problem calculate the radial and angular acceleration of the ball and apply Newton’s second law of motion to calculate the value of unknown forces

Fundamentals

Suppose a body is loaded by some external force F and the mass of the body is m, the body is accelerating with a in the direction of force, then according to Newton’s second law:

F =ma
та

Similarly, in case of rotation of body due to some external torque T, and the body is having angular acceleration of , then by Newton’s second law of motion:

T Ia

The acceleration of a particle in cylindrical coordinate system can be written as:

(F-ri)e, +(ró+2t0)e, +že,

Here, (i-ró)
is the acceleration in radial direction, -2iô
(rö+2+d)
is the acceleration in azimuth direction or angular direction and is the acceleration in z direction.

Consider the angular displacement of the ball.

= T.
…… (1)

Substitute 0.5 s for t.

0-тx(0.5)*
л
. гad
4
=

Differentiate equation (1) with respect time t.

de d(xi*
dt
dt
…… (2)

Substitute 0.5 s for t.

-2xt
=2 x0.5
%3Dя гаd/s

Again, differentiate equation (2) with respect to time t.

do d(r2t)
dt
dt
=2n

Consider the radial displacement of the ball:

r 0.6sin0
…… (3)

Substitute 4
for .

r = 0.6x sin|
4
= 0.4243 m

Differentiate equation (3) with respect to .

(0.6sin0
dr
d0
0.6cos0xÒ
…… (4)

Substitute 4
for , rad/s
for and 2x rad/s
2
for .

0.6cos
хл
- 1.332 m/s

Again, differentiate equation (4) with respect to .

di_d(0.6cos0x)
d0
d0
=0.6cos0xö+0.
-sin0xé
0.6 cos0xö-0.6(sin 0)&

Substitute 4
for , rad/s
for and 2x rad/s
2
for .

0.6cos
2
x 2t-0.6sin
=-1.5215 m/s2

Calculate the radial component of acceleration.

a, -ro

Substitute -1.5215 m/s2
for , 0.4243 m
for r and rad/s
for .

a, -1.5215-0.4243x(x
=-5.709173 m/s

Calculate the angular component of acceleration.

arö+2r0

Substitute 0.4243 m
for r, 2x rad/s
2
for , 1.332 m/s
for and for .

|a, = 0.4243x 27n +2x 1.332 x z7
11.03515 m/s

Draw the free-body diagram of the ball at t 0.5s
.

\45°
0.3m
W
45°

Apply Newton’s second law of motion in radial direction.

ΣΕ-ma
F cos0 N cos0-mg cos0 = ma

Substitute 45
for , 0.2 kg for m, 9.81 m/s2
for g and -5.709173 m/s
for а,
.

F cos 45° - Ncos 45° -0.2x9.8lcos 45° = 0.2x (-5.709173)
cos 45°(F-N)1.387-1.141
F-N 0.348
…… (5)

Apply Newton’s second law of motion in direction.

=
F sin 0N sin 0- mg sin 0 = ma,

Substitute 45
for , 0.2 kg for m, 9.81 m/s2
for g and 11.03515 m/s2
for а,
.

Fsin 45°Nsin 45° -0.2x9.81sin 45° = 0.2x(11.03515)
sin 45° (F+ N) 2.20703 +1.3873
F N 5.0832
…… (6)

Add equation (5) and (6).

2F 0.348+5.0832
2F 5.4312
F =2.7156 N

Ans:

The magnitude of force F exerted by the blower on the ball is 2.7156 N
.

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