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A merry-go-round rotates at the rate of
0.2 rev/s with an 88 kg man standing at a
point 2.1 m from the axis of rotation.
What is the new angular speed when the
man walks to a point 0 m from the center?
Consider the merry-go-round is a solid 75 kg
cylinder of radius of 2.1 m.
Answer in units of rad/s.

Sagot :

MDKPfromIDN

So, the new angular speed when the man walks to a point 0 m from the center (in rad/s) is approximately 0.87π rad/s.

Introduction

Hi ! I will help you with this problem. This problem will mostly adopt the principle of rotational dynamics, especially angular momentum. When there is a change in the mass or radius of rotation of an object, it will not affect its angular momentum, only its angular velocity will change. Therefore, the law of conservation of angular momentum can be written as :

[tex] \sf{L_1 = L_2} [/tex]

[tex] \sf{I_1 \times \omega_1 = I_2 \times \omega_2} [/tex]

[tex] \boxed{\sf{\bold{m_1 \times (r_1)^2 \times \omega_1 = m_2 \times (r_2)^2 \times \omega_2}}} [/tex]

With the following condition :

  • [tex] \sf{m_1} [/tex] = initial mass (kg)
  • [tex] \sf{m_2} [/tex] = final object mass (kg)
  • [tex] \sf{r_1} [/tex] = initial turning radius (m)
  • [tex] \sf{r_2} [/tex] = final turning radius (m)
  • [tex] \sf{\omega_1} [/tex] = initial angular velocity (rad/s)
  • [tex] \sf{\omega_2} [/tex] = final angular velocity (rad/s)

Rationale

Previously, we assumed that when rotating, only merry-go-round that experienced inertia so the man will only add to the mass of the merry-go-round. Perhaps, having a man at the end of it will change the slope (direction of torque), but not change the value of angular momentum. So, when this person is in the middle or on the central axis (read: 0 m from the center), the man no longer had any effect on the weight gain.

Problem Solving

We know that :

  • [tex] \sf{m_1} [/tex] = initial mass = 88 (man) + 75 (object) = 163 kg
  • [tex] \sf{m_2} [/tex] = final object mass = 75 kg
  • [tex] \sf{r_1} [/tex] = [tex] \sf{r_2} [/tex] = r >> The radius of rotation is always the same
  • [tex] \sf{\omega_1} [/tex] = initial angular velocity = 0.2 rev/s = 0.2 × 2π rad/s >> 0.4π rad/s

What was asked :

  • [tex] \sf{\omega_2} [/tex] = final angular velocity = ... rad/s

Step by step :

[tex] \sf{m_1 \times (r_1)^2 \times \omega_1 = m_2 \times (r_2)^2 \times \omega_2} [/tex]

[tex] \sf{163 \times \cancel{(r)^2} \times 0.4 \pi = 75 \times \cancel{(r)^2} \times \omega_2} [/tex]

[tex] \sf{65.2 \pi = 75 \times \omega_2} [/tex]

[tex] \sf{\omega_2 = \frac{65.2 \pi}{75}} [/tex]

[tex] \boxed{\sf{\omega_2 \approx 0.87 \pi \: rad/s}} [/tex]

Conclusion

So, the new angular speed when the man walks to a point 0 m from the center (in rad/s) is approximately 0.87π rad/s

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