Physics, 08.04.2020 02:54 heroxtree

A merry-go-round rotates at the rate of 0.14 rev/s with an 84 kg man standing at a point 2.4 m from the axis of rotation. What is the new angular speed when the man walks to a point 0 m from the center

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Answer from: Leigh305

The given question is incomplete. The complete question is as follows.

A merry-go-round rotates at the rate of 0.14 rev/s with an 84 kg man standing at a point 2.4 m from the axis of rotation. What is the new angular speed when the man walks to a point 0.6 m from the center? Assume that the merry-go-round is a solid 25-kg cylinder of radius 2.0 m.

Explanation:

The given data is as follows.

Initial angular velocity ( $\omega_{1}$ ) = 0.14 rev/s

$\omega_{1} = (0.14 \times 2 \pi) rad/s$

= 0.87 rad/s

Mass of man (M) = 84 kg, Mass of cylinder (m) = 25 kg

Radius of the cylinder (r) = 2.4 m

Let $r_{1}$ be the initial distance of man from the axis of rotation = 2.0 m

Final distance of man from the axis of rotation ( $r_{2}$ ) = 0.6 m

Formula to calculate the moment of inertia of cylinder is as follows.

I = $\frac{Mr^{2}}{2}$

And, moment of inertia of man = $mr^{2}$

Therefore, initial moment of inertia of the system is as follows.

$I_{1} = \frac{Mr^{2}}{2} + mr^{2}_{1}$

= $\frac{84 \times (2.4 m)^{2}}{2} + 25 \times (2)^{2}$

= 241.92 + 100

= 341.92 $kg m^{2}$

Now, final moment of inertia of the system will be calculated as follows.

$I_{2} = \frac{Mr^{2}}{2} + mr^{2}_{2}$

= $\frac{84 \times (2.4)^{2}}{2} + 25 \times (0.6)^{2}$

= 241.92 + 9

= 250.92

Now, according to the law of conservation of angular momentum is as follows.

Initial angular momentum = Final angular momentum

$I_{1} \omega_{1} = I_{2} \omega_{2}$

or, $\omega_{2} = \frac{I_{1}\omega_{1}}{I_{2}}$

= $\frac{341.92 \times 0.87}{250.92}$

= 1.18 rad/s

Thus, we can conclude that the new angular speed is 1.18 rad/s.

Answer from: Quest

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

-reverse card

Answer from: Quest

From negative to negative

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