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3 years ago

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A cylindrical rod 21.5 cm long with a mass of 1.20 kg and a radius of 1.50 cm has a ball of diameter of 6.90 cm and a mass of 2.

00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top. The apparatus is free to pivot about the bottom end of the rod.
(a) After it falls through 90°, what is itsrotational kinetic energy?
J
(b) What is the angular speed of the rod and ball?
rad/s
(c) What is the linear speed of the ball?
m/s
(d) How does this compare with the speed if the ball had fallenfreely through the same distance of 24.8 cm?
vswing is ---Select---greater thanless thanvfall by %

1 answer:

Sergeeva-Olga [200]3 years ago

6 0

Answer

given,

length of rod = 21.5 cm = 0.215 m

mass of rod (m) = 1.2 Kg

radius, r = 1.50

mass of ball, M = 2 Kg

radius of ball, r = 6.90/2 = 3.45 cm = 0.0345 m

considering the rod is thin

$I = \dfrac{1}{3}M_{rod}L^2 + [\dfrac{2}{5}M_{ball}R^2+M_{ball}(R+L)^2]$

$I = \dfrac{1}{3}\times 1.2 \times 0.215^2 + [\dfrac{2}{5}\times 2 \times 0.0345^2+2\times (0.0345 +0.215)^2]$

I = 0.144 kg.m²

rotational kinetic energy of the rod is equal to

$KE = M_{rod}g\dfrac{L}{2} + M_{ball}g(L+R)^2$

$KE = 1.2 \times 9.8 \times \dfrac{0.215}{2} + 2\times 9.8\times (0.215+0.0345)^2$

KE = 6.15 J

b) using conservation of energy

$K_f + U_f = K_i + U_i + \Delta E$

$\dfrac{1}{2}I\omega^2+ 0=0 + 6.15+0$

$\dfrac{1}{2}\times 0.144 \times \omega^2= 6.15$

ω = 9.25 rad/s

c) linear speed of the ball

v = r ω

v = (L+R )ω

v = (0.215+0.0345) x 9.25

v =2.31 m/s

d) using equation of motion

v² = u² + 2 g h

v² = 0 + 2 x 9.8 x 0.248

v = √4.86

v =2.20 m/s

speed attained by the swing is more than free fall

% greater = $\dfrac{2.31-2.20}{2.20}\times 100$

= 5 %

speed of swing is 5 % more than free fall

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b) $t=0.495 s$

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<u>x-component: </u>

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Where:

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$y=0$ is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(14.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(52\°))$ (9)

$x=19.684 m$ (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

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$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=75.99\°$

$\theta_{2}=37.99\° \approx 38\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of $t_{1}$ and $t_{2}$ from (6) and (11), respectively:

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$

$t_{1}=-\frac{2(14.1 m/s)sin(52\°)}{-9.8m/s^{2}}$ (16)

$t_{1}=2.267 s$ (17)

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$

$t_{2}=-\frac{2(14.1 m/s)sin(38\°)}{-9.8m/s^{2}}$ (18)

$t_{2}=1.771 s$ (19)

Since snowball 1 was thrown before snowball 2, we have:

$t_{1}-t=t_{2}$ (20)

Finding the time difference $t$ between both:

$t=t_{1}-t_{2}$ (21)

$t=2.267 s - 1.771 s$

Finally:

$t=0.495 s$

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