Q19. A 5-lb ball B is traveling around in a horizontal circle of radius r₁ = 4.6 ft with aspeed (VB)₁ = 4.9 ft/s. If the attached cord is pulled down through the hole with aconstant speed v₁ = 2.1 ft/s, determine how far is the ball from the hole when theball reaches a speed of 15 ft/s. Please pay attention: the numbers may change sincethey are randomized. Your answer must include 2 places after the decimal point, andproper unit.B ri(VB)1Your Answer:AnswerVrunits

Answered: Q19. A 5-lb ball B is traveling around…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Q19. A 5-lb ball B is traveling around in a horizontal circle of radius r₁ = 4.6 ft with a
speed (VB)₁ = 4.9 ft/s. If the attached cord is pulled down through the hole with a
constant speed v₁ = 2.1 ft/s, determine how far is the ball from the hole when the
ball reaches a speed of 15 ft/s. Please pay attention: the numbers may change since
they are randomized. Your answer must include 2 places after the decimal point, and
proper unit.
B ri
(VB)1
Your Answer:
Answer
Vr
units

Expert Solution

Step 1

To find-:

The final distance of ball from the hole.

Given-:

The weight of the ball is $W = 5 l b$ .

The initial radius of rotation of ball is $r_{1} = 4.61 f t$ .

The initial speed of the ball is ${(V_{B})}_{1} = 4.9 \frac{f t}{s}$ .

The speed of the chord with which it is being pulled is $V_{r} = 2.1 \frac{f t}{s}$ .

The final speed of the ball is $V_{B} = 15 \frac{f t}{s}$ .

FBD-:

Calculations-:

At given final position, the net velocity of the ball is the resultant of final tangential velocity of ball ${(V_{B})}_{2}$ and velocity of cord $V_{r}$ i.e-:

$\vec{V_{B}} = {(\vec{V_{B}})}_{2} + \vec{V_{r}}$

${(V_{B})}^{2} = {{(V_{B})}_{2}}^{2} + {(V_{r})}^{2}$ (1)

Substitute the value of $V_{B} = 15 \frac{f t}{s}$ and $V_{r} = 2.1 \frac{f t}{s}$ in equation (1).

$\begin{array}{rcl} {(15)}^{2} & = & {{(V_{B})}_{2}}^{2} + {(2.1)}^{2} \\ {{(V_{B})}_{2}}^{2} & = & 225 - 4.41 \\ {(V_{B})}_{2} & = & 14.85 \frac{f t}{s} \end{array}$

Since, there is no external torque, hence angular momentum will remains conserved. So, initial angular momentum is equal to final angular momentum.

$m {(V_{B})}_{1} r_{1} = m {(V_{B})}_{2} r_{2}$

Here, $m$ is the mass of the ball and $r_{2}$ is the final distance of ball from the hole.

${(V_{B})}_{1} r_{1} = {(V_{B})}_{2} r_{2}$ (2)

Substitute the value of ${(V_{B})}_{1} = 4.9 \frac{f t}{s}$ , $r_{1} = 4.61 f t$ and ${(V_{B})}_{2} = 14.85 \frac{f t}{s}$ in equation (2).

$4.9 \times 4.61 = 14.85 \times r_{2} r_{2} = 1.52 f t$

Thus, the final distance of ball from the hole is $r_{2} = 1.52 f t$ .

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