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3 years ago

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A playground merry-go-round of radius R = 1.40 m has a moment of inertia I = 265 kg · m2 and is rotating at 11.0 rev/min about a

frictionless vertical axle. Facing the axle, a 27.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?

1 answer:

Tom [10]3 years ago

6 0

Answer:

The value of new value of angular speed of merry go round. $\omega_{2}$ = 0.96 $\frac{rad}{sec}$

Explanation:

Given data

r = 1.4 m

Moment of inertia $I_{1}$ = 265 kg - $m^{2}$

$N_{1} =$ 11 RPM

$\omega_{1} = \frac{2 \pi N}{60}$

$\omega_{1} = \frac{2 \pi (11)}{60}$

$\omega_{1}$ = 1.15 $\frac{rad}{sec}$

From conservation of momentum principal

$I_{1} \omega_{1} = I_{2} \omega_{2}$ ------- (1)

$I_{2} = m r^{2} + 265$

$I_{2} = 27 (1.4)^{2} + 265$

$I_{2} = 317.92 \ kg m^{2}$

Put all the values in equation (1)

265 × 1.15 = 317.92 × $\omega_{2}$

$\omega_{2}$ = 0.96 $\frac{rad}{sec}$

This is the value of new value of angular speed of merry go round.

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Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and

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The magnitude of the average friction force exerted on the collar (F)= 8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

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Here we are given,

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Now we put the known values in the above equation, we get;

$\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c} = mgy$

Or, $\frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2} = 3.55\times 9.81\times 1.80$

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)= 8.641 N

Learn more about friction:

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Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .

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