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Shkiper50 [21]
3 years ago
15

A lumberjack (mass = 103 kg) is standing at rest on one end of a floating log (mass = 277 kg) that is also at rest. The lumberja

ck runs to the other end of the log, attaining a velocity of +3.74 m/s relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water. (a) What is the velocity of the first log (again relative to the shore) just before the lumberjack jumps off? (b) Determine the velocity of the second log (again relative to the shore) if the lumberjack comes to rest relative to the second log.
Physics
1 answer:
Lorico [155]3 years ago
5 0

Answer:

Explanation:

Given

mass of lumber jack m_L=1000 kg

mass of floating log m_f=277 kg

velocity of lumber jack =3.74 m/s (w.r.t. shore)

conserving momentum

Initial momentum=Final momentum

0=m_L\times 3.74+m_f\times v

v=-\frac{1000\times 3.74}{277}

v=-13.501 m/s

(b)For second Log

Conserving Momentum

Initial momentum of log and Lumber jack=Final momentum of log and lumberjack

m_L\times 3.74=(m_L+m_f)u

u is final velocity of lumberjack and log

u=\frac{m_L}{m_L+m_f}\times 3.74

u=\frac{1000}{1000+277}=2.92 m/s

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At t = 4.2 s

Angular velocity: 6. 17 rad /s

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There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

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therefore, equating this to the above equation gives

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Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

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

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

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

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

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Therefore, ar t = 4.2 s, the above gives

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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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