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3 years ago

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A 780 g , 54-cm-long metal rod is free to rotate about a frictionless axle at one end. While at rest, the rod is given a short b

ut sharp 1000 N hammer blow at the center of the rod, aimed in a direction that causes the rod to rotate on the axle. The blow lasts a mere 2.5 ms .What is the rod's angular velocity immediately after the blow?

1 answer:

evablogger [386]3 years ago

6 0

Answer:

The angular velocity is <em>35.5 rad/s.</em>

Explanation:

Step 1: Find the moment of Inertia for the rod

The moment of inertia, <em>I</em>, of a rod:

<em> $I=\frac{1}{12} ML^{2}$ .......... (1)</em>

where M = mass of rod (0.78 Kg); L = Length of rod (0.54 m).

$I=\frac{1}{12} (0.78)(0.54)^{2} =0.019\frac{Kg}{m^{2} }$

Step 2: Calculate the angular acceleration from Rotational kinetic notation

<em>F.r = I.α ........... (2)</em>

<em>where F is the force acting upon the rod; r is the half length of the rod; I is the moment of Inertia and; α is the angular acceleration.</em>

<em>∴ (1000 N)(0.27 m) = 0.019α</em>

<em>α = 270 Nm / 0.019 Kgm²</em>

<em>α = 14210.5 rad/s</em>

Step 3: We find the angular velocity by using the equation below:

<em>ωf = ωi + αt ......... (3)</em>

<em>where </em>

<em>ωf is the angular velocity after the blow</em>

<em>ωi is the angular velocity before the blow = 0</em>

<em>t is the time taken for the blow to occur = 2.5 ms</em>

<em>ωf = 0 + (14210.5 rad/s)(2.5 ms) = 35.5 rad/s.</em>

<em />

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$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

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The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

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Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

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