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2 years ago

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b) Assume the rod is 0.60 m long and has a mass of 0.50 kg, and the clay blob has a mass of 0.20 kg and moves at an initial velo

city of 8.0 m/s. Calculate the final angular velocity of the rod. Be sure to put units in your calculation and show the resulting units in your answer.

1 answer:

SOVA2 [1]2 years ago

4 0

Answer:

The correct answer is "6.96 rad/s".

Explanation:

The given values are:

Length,

L = 0.6 m

Mass,

m₁ = 0.5 kg

m₂ = 0.2 kg

Initial velocity,

V = 8 m/s

Now,

The final angular velocity will be:

⇒ $\omega =\frac{6m_1V}{(4m_1+3m_2)L}$

By substituting the values, we get

⇒ $=\frac{6\times 0.2\times 8}{(4\times 0.2+3\times 0.5)0.6}$

⇒ $=\frac{9.6}{1.38}$

⇒ $=6.96 \ rad/s$

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3 years ago

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

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Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

<u>Using the equation of motion:</u>

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

<u>Now the new velocity:</u>

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

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