Answered

Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

A merry-go-round rotates at the rate of 0.17 rev/s with an 79 kg man standing at a point 1.6 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 34 kg cylinder of radius of 1.6 m.
Answer in units of rad/s.

Sagot :

leena

Hi there!

We can use the conservation of angular momentum to solve.

[tex]L_i = L_f\\\\I\omega_i = I\omega_f[/tex]

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

Recall the following equations for the moment of inertia.

[tex]\text{Solid cylinder:} I = \frac{1}{2}MR^2\\\\\text{Object around center:} = MR^2[/tex]

Begin by converting rev/sec to rad sec:


[tex]\frac{0.17rev}{s} * \frac{2\pi rad}{1 rev} = 1.068 \frac{rad}{s}[/tex]

According to the above and the given information, we can write an equation and solve for ωf.

[tex]1.068(\frac{1}{2}(34)(1.6)^2 + (79)(1.6)^2) = \omega_f(\frac{1}{2}(34)(1.6^2) + 79(0^2))\\\\\omega_f = \boxed{6.03 \frac{rad}{sec}}[/tex]

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.