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3 years ago

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A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp

ed around the disk is pulled with a force F equal to 20 N.
(a) What is the torque being exerted by this force about the rotation axis?

(b) What is the angular acceleration of the disk?

(c) If the disk starts from rest, what is its angular speed after 4.6 s?

(d) What is its kinetic energy after the 4.6 s?

(e) What is the angular displacement of the disk during the 4.6 s?

1 answer:

AlekseyPX3 years ago

3 0

Answer:

(A) 2.4 N-m

(B) $0.035kgm^2$

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque $\tau =Fr$ , here F is force and r is distance

So $\tau =20\times 0.12=2.4Nm$

(B) Moment of inertia is equal to $I=\frac{1}{2}mr^2$

So $I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2$

Torque is equal to $\tau =I\alpha$

So angular acceleration $\alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2$

(C) As the disk starts from rest

So initial angular speed $\omega _{0}=0rad/sec$

Time t = 4.6 sec

From first equation of motion we know that $\omega =\omega _0+\alpha t$

So $\omega =0+68.571\times 4.6=315.426rad/sec$

(D) Kinetic energy is equal to $KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J$

(E) From second equation of motion

$\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad$

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Friction force N₁μ pushing parallel up AB,

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Weight force P pulling down,

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Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

$d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at$

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

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Treat them separately.

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First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

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Vi = ? Θ =?

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We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

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$vf_y=vi_y+at$

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$0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s$

So we have our first initial vertical velocity:

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T= 2t

T = 2 (0.782s)

<em>T = 1.564s</em>

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<em> $d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x$ </em>

Vix=6.39m/s

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c²=a²+b²

Using it in the context of our problem:

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Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

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