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3 years ago

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A jogger runs by a river with a velocity of 7 m s relative to the ground. A leaf floating on the river moves with a velocity of

2 m s relative to the ground (riverbed). What is the velocity of the leaf relative to the jogger?

1 answer:

Hunter-Best [27]3 years ago

3 0

Answer:

The velocity of the leaf relative to the jogger is 5 m/s.

Explanation:

Given that,

Velocity of jogger wrt to the ground, $V_j=7\ m/s$

velocity of leaf wrt the ground, $v_i=2\ m/s$

We need to find the velocity of the leaf relative to the jogger. Let it is equal to V. So, it is given by :

$V=v_j-v_i\\\\V=7-2\\\\V=5\ m/s$

So, the velocity of the leaf relative to the jogger is 5 m/s. Hence, this is the required solution.

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Given that,

Mass of disk = 1.2 kg

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Radius of rod = 0.11 m

Mass of small disk = 0.5 kg

Force = 29 N

Time t = 0.022 s

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Distance d= 0.039 m

(I). We need to calculate the speed of the apparatus

Using work energy theorem

$W=\Delta K.E$

$Fd=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2Fd}{M+4m}}$

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$v=0.840\ m/s$

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Using formula of torque

$\tau=I\alpha$

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$29\times0.07=(\dfrac{1}{2}\times1.2\times0.07^2+4\times0.5\times0.11^2)\alpha$

$\alpha=\dfrac{29\times0.07}{0.02714}$

$\alpha=74.79\ rad/s^2$

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$\omega=0+74.79\times0.022$

$\omega=1.645\ rad/s$

(c). We need to calculate the angular speed of the apparatus

Using equation of angular motion

$\omega_{0}^2=\omega^2+2\alpha t$

Put the value into the formula

$\omega_{0}^2=1.645^2+2\times74.79\times0.022$

$\omega=2.44\ rad/s$

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