Question

A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs tangent to the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John’s mass is 30 kg. What is the merry-goround’s angular velocity, in rpm, after John jumps on

Answer 1

Answer:

f = 22.3 rpm

Explanation:

Here we know that there is no external torque on the system of disc + John

So here we can use angular momentum conservation

So we will have

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

so here we have

$I = \frac{1}{2}MR^2$

$I = \frac{1}{2}(250)(1.5^2)$

$I = 281.25 kg m^2$

now we have

$281.25(2\pi \frac{20}{60}) + (30 \times 5 \times 1.5) = (281.25 + 30(1.5^2))\omega_f$

$589 + 225 = (281.25 + 67.5)\omega_f$

$\omega_f = 2.33$

$2\pi f = 2.33$

$f = \frac{2.33}{2\pi}$

$f = 22.3 rpm$

Answer 2

The angular velocity of the merry-go-round, in rpm, after John jumps on is obtained using the concept of angular momentum conservation and its value is 16.77 rpm.

Law of Conservation of Angular Momentum

According to the law of conservation of angular momentum, we have;

$\frac{1}{2} MR^2\,\omega_1 +mvR=\frac{1}{2} MR^2\,\omega_2+mR^2\,\omega_2$

Given values are;

M = 250kg

m = 30kg

R = 1.5 m

$\omega_1 = 30\,rpm$

v = 5 m/s

Substituting the given values, we get;

$\frac{1}{2} (250\times 1.5^2\times20\,rpm) +(30\times 5\times 1.5\,)=[\frac{1}{2}(250kg \times 1.5^2)+(30\times 1.5^2)]\omega_2\\\\(5625+225)=(281.25+67.5)\omega_2\\\\\implies \omega_2=16.77\,rpm$

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