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Sergeeva-Olga [200]

2 years ago

6

A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round, which can be modeled as a disk with

a mass of 300 kg , is spinning at 23 rpm. John runs tangent to the merry-go-round at 4.4 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg.
Required:
What is the merry-go-round's angular velocity, in rpm, after John jumps on?

1 answer:

AveGali [126]2 years ago

4 0

Answer:

The merry-go-round's angular velocity 23.84 RPM

Explanation:

Given;

diameter of merry go round, d = 3 m

radius of the merry go round, R = 1.5 m

mass of the merry go round, m = 300 kg

angular velocity = 23 rpm

velocity of John, v = 4.4 m/s

mass of John, m = 30 kg

Apply conservation of angular momentum;

$L_i = L_f$

$I \omega_i + mvR = (I + mR^2)\omega _f$

where;

I is moment of inertia of disk

$I = \frac{1}{2} mR^2\\\\I = \frac{1}{2} *300*1.5^2\\\\I = 337.5 \ kg.m^2$

Substitute in this value in the above equation;

$337.5(2\pi \frac{23}{60} ) + (30*4.4*1.5) = (337.5 + 30*1.5^2) \omega_f\\\\812.9925 \ + \ 198 = 405 \omega _f\\\\1010.9925 = 405 \omega _f\\\\\omega _f = \frac{1010.9925}{405} \\\\\omega _f = 2.496 \ rad/s$

1 rad/s = 9.5493 rpm

2.496 rad/s = 23.84 RPM

Therefore, the merry-go-round's angular velocity 23.84 RPM

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